Arithmetic Progression

Here, T₂ - T₁ = 15 - 9 = 6

T₃ - T₂ = 21 - 15 = 6

T₄ - T₃ = 27 - 21 = 6

Since the difference between each consecutive term is same, ∴ the progression is an AP.

So, first term = 9

Common difference = 15 - 9 = 6

Next term = T₅ = T₄ + d = 27 + 6 = 33

Here, T₂ - T₁ = 6 - 11 = - 5

T₃ - T₂ = 1 - 6 = - 5

T₄ - T₃ = - 4 - 1 = - 5

Since the difference between each consecutive term is same, ∴ the progression is an AP.

So, first term = 11

Common difference = 6 - 11 = - 5

Next term = T₅ = T₄ + d = - 4 + (-5) = - 9

Here, T₂ - T₁ = (-5/6) - (-1) = 1/6

T₃ - T₂ = (-2/3) - (-5/6) = 1/6

T₄ - T₃ = (-1/2) - (-2/3) = 1/6

Since the difference between each consecutive term is same, ∴ the progression is an AP.

So, first term = - 1

Common difference = (-5/6) - (-1) = 1/6

Next term = T₅ = T₄ + d

= (-1/2) + (1/6)

= (-2/6)

= (-1/3)

Here, T₂ - T₁ = √8 - √2

= 2√2 - √2

= √2

T₃ - T₂ = = √18 - √8

= 3√2 - 2√2

= √2

T₄ - T₃ = = √32 - √18

= 4√2 - 3√2

= √2

Since the difference between each consecutive term is same, ∴ the progression is an AP.

So, first term = √2

Common difference = √8 - √2 = √2

Next term = T₅ = T₄ + d

= √32 + √2

= 4√2 + √2

= 5√2

= √50

Here, T₂ - T₁ = √45 - √20

= 3√5 - 2√5

= √5

T₃ - T₂ = = √80 - √45

= 4√5 - 3√5

= √5

T₄ - T₃ = = √125 - √80

= 5√5 - 4√5

= √5

Since the difference between each consecutive term is same, ∴ the progression is an AP.

So, first term = √20

Common difference = √45 - √20 = √5

Next term = T₅ = T₄ + d

= √125 + √5

= 5√5 + √5

= 6√5

= √180

Here, First term = a = 9

Common difference = d = 13 - 9 = 4

To find = 20^th term, ∴ n = 20

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = 9 + (20 - 1) × 4

⇒ a_n = 9 + 19 × 4 = 9 + 76 = 85

∴ 20^th term of the given AP is 85.

Here, First term = a = 20

Common difference = d = 17 - 20 = - 3

To find = 35^th term, ∴ n = 35

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = 20 + (35 - 1) × (-3)

⇒ a_n = 20 + 34 × (-3) = 20 - 102 = - 82

∴ 20^th term of the given AP is - 82.

The given AP can be rewritten as √2, 3√2, 5√2, 7√2….

Here, First term = a = √2

Common difference = d = 3√2 - √2 = 2√2

To find = 18^th term, ∴ n = 18

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = √2 + (18 - 1) × 2√2

⇒ a_n = √2 + 17 × 2√2 = √2 + 34√2 = 35√2

∴ 18^th term of the given AP is 35√2.

Here, First term = a = 3/4

Common difference = d = 5/4 - 3/4 = 2/4

To find = 9^th term, ∴ n = 9

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = (3/4) + (9 - 1) × (2/4)

⇒ a_n = 3/4 + 8 × (2/4) = 3/4 + 16/4 = 19/4

∴ 9^th term of the given AP is 19/4.

Here, First term = a = - 40

Common difference = d = - 15 - (-40) = 25

To find = 15^th term, ∴ n = 15

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = - 40 + (15 - 1) × (25)

⇒ a_n = - 40 + 14 × (25) = - 40 + 350 = 310

∴ 15^th term of the given AP is 310.

The given AP can be rewritten as 6, 31/4 ,19/2 , 45/4,…

Here, First term = a = 6

Common difference = d = (31/4) - 6 = 7/4

To find = 37^th term, ∴ n = 37

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = 6 + (37 - 1) × (7/4)

⇒ a_n = 6 + 36 × (7/4) = 6 + 63 = 69

∴ 37^th term of the given AP is 69.

Here, First term = a = 5

Common difference = d = 9/2 - 5 = - (1/2)

To find = 25^th term, ∴ n = 25

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = 5 + (25 - 1) × (-1/2)

⇒ a_n = 5 + 24 × (-1/2) = 5 - 12 = - 7

∴ 25^th term of the given AP is - 7.

Here, First term = a = 5

Common difference = d = 11 - 5 = 6

To find = n^th term

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = 5 + (n - 1) × 6

⇒ a_n = 5 + 6n - 6 = 6n - 1

∴ n^th term of the given AP is (6n - 1).

Here, First term = a = 16

Common difference = d = 9 - 16 = - 7

To find = n^th term

Using the formula for finding n^th term of an A.P.,

a_n = a + (n - 1) × d

∴ a_n = 16 + (n - 1) × (-7)

⇒ a_n = 16 - 7n + 7 = 23 - 7n

∴ n^th term of the given AP is (23 - 7n).

n^th term of the AP is (4n - 10).

For n = 1, we have a₁ = 4 - 10 = - 6

For n = 2, we have a₂ = 8 - 10 = - 2

For n = 3, we have a₃ = 12 - 10 = 2

For n = 4, we have a₄ = 16 - 10 = 6, and so on.

∴ a₄ - a₃ = a₃ - a₂ = a₂ - a₁ = 4 = constant.

∴ the given progression is an AP.

Hence, (i) Its first term = a = - 6

(ii) common difference = 4

(iii) To find :16^th term

∴ a₁₆ = a + (16 - 1)d

⇒ a₁₆ = - 6 + 15 × 4 = 54

∴ 16^th term of the given AP is 54.

In the given AP, the first term = a = 6

Common difference = d = 10 - 6 = 4

Last term = 174

To find: No. of terms in the AP.

Since, we know that

a_n = a + (n - 1) × d

∴ 174 = 6 + (n - 1) × 4

⇒ 174 - 6 = 4n - 4

⇒ 168 = 4n - 4

⇒ 168 + 4= 4n

⇒ 4n = 172

⇒ n = 172/4

⇒ n = 43

∴ Number of terms = 43.

In the given AP, the first term = a =41

Common difference = d = 38 - 41 = - 3

Last term = 8

To find: No. of terms in the AP.

Since, we know that

a_n = a + (n - 1) × d

∴ 8 = 41 + (n - 1) × (-3)

⇒ 8 - 41 = - 3n + 3

⇒ - 33 = - 3n + 3

⇒ - 33 - 3 = - 3n

⇒ - 3n = - 36

⇒ n = 36/3

⇒ n = 12

∴ Number of terms = 12.

In the given AP, the first term = a = 18

Common difference = d = (31/2) - 18 = (-5/2)

Last term = - 47

To find: No. of terms in the AP.

Since, we know that

a_n = a + (n - 1) × d

∴ - 47 = 18 + (n - 1) × (-5/2)

⇒ - 47 - 18= (n - 1) × (-5/2)

⇒ - 65 = (n - 1) × (-5/2)

⇒ - 65 × (-2/5) = n - 1

⇒ n - 1 = 26

⇒ n = 26 + 1

⇒ n = 27

∴ Number of terms = 27.

In the given AP, the first term = a = 3

Common difference = d = 8 - 3 = 5

To find: place of the term 88.

So, let a_n = 88

Since, we know that

a_n = a + (n - 1) × d

∴ 88 = 3 + (n - 1) × 5

⇒ 88 - 3 = 5n - 5

⇒ 85 = 5n - 5

⇒ 85 + 5 = 5n

⇒ 5n = 90

⇒ n = 90/5

⇒ n = 18

∴ 18^th term of the AP is 88.

In the given AP, the first term = a = 72

Common difference = d = 68 - 72 = - 4

To find: place of the term 0.

So, let a_n = 0

Since, we know that

a_n = a + (n - 1) × d

∴ 0 = 72 + (n - 1) × (-4)

⇒ 0 - 72 = - 4n + 4

⇒ - 72 - 4 = - 4n

⇒ - 76 = - 4n

⇒ n = 76/4

⇒ n = 19

∴ 19^th term of the AP is 0.

In the given AP, the first term = a = 5/6

Common difference = d = 1 - 5/6 = 1/6

To find: place of the term 3.

So, let a_n = 3

Since, we know that

a_n = a + (n - 1) × d

∴ 3 = (5/6) + (n - 1) × (1/6)

⇒ 3 - (5/6) = (n - 1) × (1/6)

⇒ 13/6 = (n - 1) × (1/6)

⇒ 13 = n - 1

⇒ n = 13 + 1

⇒ n = 14

∴ 14^th term of the AP is 3.

In the given AP, the first term = a = 21

Common difference = d = 18 - 21 = - 3

To find: place of the term - 81.

So, let a_n = - 81

Since, we know that

a_n = a + (n - 1) × d

∴ - 81 = 21 + (n - 1) × (-3)

⇒ - 81 - 21 = - 3n + 3

⇒ - 102 = - 3n + 3

⇒ - 102 - 3 = - 3n

⇒ - 3n = - 105

⇒ n = 105/3

⇒ n = 35

∴ 35^th term of the AP is - 81.

In the given AP, the first term = a = 3

Common difference = d = 8 - 3 = 5

To find: place of the term which is 55 more than its 20^th term.

So, we first find its 20^th term.

Since, we know that

a_n = a + (n - 1) × d

∴ a₂₀ = 3 + (20 - 1) × 5

⇒ a₂₀ = 3 + 19 × 5

⇒ a₂₀ = 3 + 95

⇒ a₂₀ = 98

∴ 20^th term of the AP is 98.

Now, 55 more than 20^th term of the AP is 55 + 98 = 153.

So, to find: place of the term 153.

So, let a_n = 153

Since, we know that

a_n = a + (n - 1) × d

∴ 153 = 3 + (n - 1) × 5

⇒ 153 - 3 = 5n - 5

⇒ 150 = 5n - 5

⇒ 150 + 5 = 5n

⇒ 5n = 155

⇒ n = 155/5 = 31

∴ 31^st term of the AP is the term which is 55 more than 20^th term.

In the given AP, the first term = a = 5

Common difference = d = 15 - 5 = 10

To find: place of the term which is 130 more than its 31^st term.

So, we first find its 31^st term.

Since, we know that

a_n = a + (n - 1) × d

∴ a₃₁ = 5 + (31 - 1) × 10

⇒ a₃₁ = 5 + 30 × 10

⇒ a₃₁ = 5 + 300

⇒ a₃₁ = 305.

∴ 31^st term of the AP is 305.

Now, 130 more than 31^st term of the AP is 130 + 305 = 435.

So, to find: place of the term 435.

So, let a_n = 435

Since, we know that

a_n = a + (n - 1) × d

∴ 435 = 5 + (n - 1) × 10

⇒ 435 - 5 = 10n - 10

⇒ 430 = 10n - 10

⇒ 430 + 10 = 10n

⇒ 10n = 440

⇒ n = 440/10 = 44

∴ 44^th term of the AP is the term which is 130 more than 31^st term.

Given: 10^th term of the AP is 52.

17^th term is 20 more than the 13^th term.

Let the first term be a and the common difference be d.

Since,

a_n = a + (n - 1) × d

therefore for 10^th term, we have,

52 = a + (10 - 1) × d

⇒ 52 = a + 9d ………… (1)

Now, 17^th term is 20 more than the 13^th term.

∴ a₁₇ = 20 + a₁₃

⇒ a + (17 - 1)d = 20 + a + (13 - 1)d

⇒ 16d = 20 + 12d

⇒ 4d = 20

⇒ d= 5

∴ from equation (1), we have,

52 = a + 9d

⇒ 52 = a + 9 × 5

⇒ 52 = a + 45

⇒ a = 52 - 45

⇒ a = 7

∴ AP is a, a + d, a + 2d, a + 3d,…

∴ AP is 7, 12, 17, 22….

First term of the AP = 6

Common difference = d = 13 - 6 = 7

Last term = 216

Since

a_n = a + (n - 1) × d

∴ 216 = 6 + (n - 1) × 7

⇒ 216 - 6 = 7n - 7

⇒ 210 = 7n - 7

⇒ 210 + 7 = 7n

⇒ 7n = 217

⇒ n = 217/7 = 31

∴ Middle term is (31 + 1)/2 = 16^th

So, a₁₆ = a + (16 - 1) × d

∴ a₁₆ = 6 + 15 × 7

⇒ a₁₆ = 6 + 105 = 111

∴ Middle term of the AP is 111.

First term of the AP = 10

Common difference = d = 7 - 10 = - 3

Last term = - 62

Since

a_n = a + (n - 1) × d

∴ - 62 = 10 + (n - 1) × (-3)

⇒ - 62 - 10 = - 3n + 3

⇒ - 72 = - 3n + 3

⇒ - 72 - 3 = - 3n

⇒ 3n = 75

⇒ n = 75/3 = 25

∴ Middle term is (25 + 1)/2 = 13^th

So, a₁₃ = a + (13 - 1) × d

∴ a₁₃ = 10 + 12 × (-3)

⇒ a₁₃ = 10 - 36 = - 26

∴ Middle term of the AP is - 26.

First term of the AP = - (4/3)

Common difference = d = - 1 - (-4/3) = - 1 + (4/3) = 1/3

Last term = 13/3

Since

a_n = a + (n - 1) × d

∴ 13/3 = (-4/3) + (n - 1) × (1/3)

⇒ (13/3) + (4/3) = (n - 1) × (1/3)

⇒ 17/3 = (n - 1) × (1/3)

⇒ 17 = n - 1

⇒ n = 17 + 1

⇒ n = 18

∴ Two middle most terms of the AP are 18/2 and (18/2) + 1 terms, i.e. 9^th and 10^th terms.

So, a₉ = a + (9 - 1) × d

∴ a₉ = (-4/3) + [8 × (1/3)]

⇒ a₉ = (-4/3) + (8/3) = 4/3

Also, a₁₀ = a₉ + d

= (4/3) + (1/3)

= 5/3

Now, a₁₀ + a₉ = (4/3) + (5/3)

= 9/3

= 3

∴ Sum of two middle most terms of the AP is 3.

Here, First term = a = 7

Common difference = d = 10 - 7 = 3

Last term = l = 184

To find: 8^th term from end.

So, n^th term from end is given by:

a_n = l - (n - 1)d

∴ 8^th term from end is:

a₈ = 184 - (8 - 1) × 3

= 184 - 21

= 163

Here, First term = a = 17

Common difference = d = 14 - 17 = - 3

Last term = l = - 40

To find: 6^th term from end.

So, n^th term from end is given by:

a_n = l - (n - 1)d

∴ 6^th term from end is:

a₆ = - 40 - (6 - 1) × (-3)

= - 40 + 15

= - 25

Here, First term = a = 3

Common difference = d = 7 - 3 = 4

Now, to check: 184 is a term of the AP or not.

Since, n^th term of an AP is given by:

a_n = a + (n - 1)d

If 184 is a term of the AP, then it must satisfy this equation.

So, let a_n = 184

∴ 184 = 3 + (n - 1) × 4

⇒ 184 - 3 = 4n - 4

⇒ 181 = 4n - 4

⇒ 181 + 4 = 4n

⇒ 4n = 185

⇒ n = 185/4 = 46.25

But this is not possible because n is the number of terms which can’t be a fraction.

Therefore, 184 is not a term of the given AP.

Here, First term = a = 11

Common difference = d = 8 - 11 = - 3

Now, to check: - 150 is a term of the AP or not.

Since, n^th term of an AP is given by:

a_n = a + (n - 1)d

If - 150 is a term of the AP, then it must satisfy this equation.

So, let a_n = - 150

∴ - 150 = 11 + (n - 1) × (-3)

⇒ - 150 - 11 = - 3n + 3

⇒ - 161 = - 3n + 3

⇒ - 161 - 3 = - 3n

⇒ 3n = 164

⇒ n = 164/3 = 54.66

But this is not possible because n is the number of terms which can’t be a fraction.

Therefore, - 150 is not a term of the given AP.

Here, First term = a = 121

Common difference = d = 117 - 121 = - 4

Let n^th term of the AP be its first negative term.

∴ a_n <0

Since, n^th term of an AP is given by:

a_n = a + (n - 1)d

∴ a + (n - 1)d < 0

⇒ 121 + (n - 1) × (-4) < 0

⇒ - 4n + 125 < 0

⇒ - 4n < - 125

⇒ 4n > 125

⇒ n > 31.25

Since n is an integer, therefore n must be 32.

∴ 32^nd term will be the first negative term of the AP.

Here, First term = a = 20

Common difference = d = (77/4) - 20 = (-3/4)

Let n^th term of the AP be its first negative term.

∴ a_n <0

Since, n^th term of an AP is given by:

a_n = a + (n - 1)d

∴ a + (n - 1)d < 0

⇒ 20 + (n - 1) × (-3/4) < 0

⇒ 80 + (n - 1) × (-3) < 0 (multiplying both sides by 4)

⇒ 80 - 3n + 3 < 0

⇒ - 3n < - 83

⇒ 3n > 83

⇒ n > 27.66

Since n is an integer, therefore n must be 28.

∴ 28^th term will be the first negative term of the AP.

Let a be the first term and d be the common difference.

Given: a₇ = - 4

a₁₃ = - 16

Now, Consider a₇ = - 4

⇒ a + (7 - 1)d = - 4

⇒ a + 6d = - 4 ……….(1)

Consider a₁₃ = - 16

⇒ a + (13 - 1)d = - 16

⇒ a + 12d = - 16 ……….(2)

Now, subtracting equation (1) from (2), we get,

6d = - 12

⇒ d = - 2

∴ from equation (1), we get,

a = - 4 - 6d

⇒ a = - 4 - 6 × (-2)

⇒ a = - 4 + 12

⇒ a = 8

Thus the AP is a, a + d, a + 2d, a + 3d, a + 4d,….

Therefore the AP is 8, 6, 4, 2, 0,….

Let a be the first term and d be the common difference.

Given: a₄ = 0

To prove: a₂₅ = 3 × a₁₁

Now, Consider a₄ = 0

⇒ a + (4 - 1)d = 0

⇒ a + 3d = 0

⇒ a = - 3d ………. (1)

Consider a₂₅ = a + (25 - 1)d

⇒ a₂₅ = - 3d + 24d (from equation (1))

⇒ a₂₅ = 21d ………. (2)

Now, consider a₁₁ = a + (11 - 1)d

⇒ a₁₁ = - 3d + 10d (from equation (1))

⇒ a₁₁ = 7d ……….(3)

From equation (2) and (3), we get,

a₂₅ = 3 × a₁₁

Hence, proved.

Let a be the first term and d be the common difference.

Given: a₈ = 0

To prove: a₃₈ = 3 × a₁₈

Now, Consider a₈ = 0

⇒ a + (8 - 1)d = 0

⇒ a + 7d = 0

⇒ a = - 7d ……. (1)

Consider a₃₈ = a + (38 - 1)d

⇒ a₃₈ = - 7d + 37d (from equation (1))

⇒ a₃₈ = 30d ………. (2)

Now, consider a₁₈ = a + (18 - 1)d

⇒ a₁₈ = - 7d + 17d (from equation (1))

⇒ a₁₈ = 10d ……….(3)

From equation (2) and (3), we get,

a₃₈ = 3 × a₁₈

Hence, proved.

Let a be the first term and d be the common difference.

Given: a₄ = 11

a₅ + a₇ = 34

To find: common difference = d

Now, Consider a₄ = 11

⇒ a + (4 - 1)d = 11

⇒ a + 3d = 11 ………….(1)

Consider a₅ + a₇ = 34

⇒ a + (5 - 1)d + a + (7 - 1)d = 34

⇒ 2a + 10d = 34

⇒ a + 5d = 17 ……..(2)

Subtracting equation (1) from equation (2), we get,

2d = 6

⇒ d = 3

∴ Common difference = d = 3

Let a be the first term and d be the common difference.

Given: a₉ = - 32

a₁₁ + a₁₃ = - 94

To find: common difference = d

Now, Consider a₉ = - 32

⇒ a + (9 - 1)d = - 32

⇒ a + 8d = - 32 ……….(1)

Consider a₁₁ + a₁₃ = - 94

⇒ a + (11 - 1)d + a + (13 - 1)d = - 94

⇒ 2a + 22d = - 94

⇒ a + 11d = - 47 ……..(2)

Subtracting equation (1) from equation (2), we get,

3d = - 15

⇒ d = - 5

∴ Common difference = d = - 5

Let a be the first term and d be the common difference.

Given: a₇ = - 1

a₁₆ = 17

Now, Consider a₇ = - 1

⇒ a + (7 - 1)d = - 1

⇒ a + 6d = - 1 ……….(1)

Consider a₁₆ = 17

⇒ a + (16 - 1)d = 17

⇒ a + 15d = 17 ……….(2)

Now, subtracting equation (1) from (2), we get,

9d = 18

⇒ d = 2

∴ from equation (1), we get,

a = - 1 - 6d

⇒ a = - 1 - 6 × (2)

⇒ a = - 1 - 12

⇒ a = - 13

Now, the n^th term of the AP is given by:

a_n = a + (n - 1)d

∴ a_n = - 13 + (n - 1) × 2

⇒ a_n = 2n - 15

∴ n^th term of the AP is (2n - 15)

Given: 4 × a₄ = 18 × a₁₈

To find : a₂₂

Consider 4 × a₄ = 18 × a₁₈

⇒ 4 [a + (4 - 1)d] = 18 [a + (18 - 1)d]

⇒ 4a + 12d = 18a + 306d

⇒ - 14 a = 294 d

⇒ a = - 21d ………………(1)

Now, a₂₂ = a + (22 - 1)d

⇒ a₂₂ = a + 21d

⇒ a₂₂ = - 21d + 21d (from equation 1)

⇒ a₂₂ = 0

∴ a₂₂ = 0

Given: 10 × a₁₀ = 15 × a₁₅

To show : a₂₅ = 0

Consider 10 × a₁₀ = 15 × a₁₅

⇒ 10 [a + (10 - 1)d] = 15 [a + (15 - 1)d]

⇒ 10a + 90d = 15a + 210d

⇒ - 5 a = 120 d

⇒ a = - 24d ………………(1)

Now, a₂₅ = a + (25 - 1)d

⇒ a₂₅ = a + 24d

⇒ a₂₅ = - 24d + 24d (from equation 1)

⇒ a₂₅ = 0

Hence, proved.

Let a be the first term and d be the common difference of the AP.

Given: a = 5

Sum of first four terms = 1/2(sum of next four terms)

⇒ a + (a + d) + (a + 2d) + (a + 3d) = 1/2 ((a + 4d) + (a + 5d) + (a + 6d) +

(a + 7d))

⇒ 4a + 6d = 1/2(4a + 22d)

⇒ 4a + 6d = 2a + 11d

⇒ 2a = 5d

⇒ d = 2a/5

As a = 5, therefore,

d = 10/5 = 2

Thus, Common difference = d = 2

Let a be the first term and d be the common difference of the AP.

Given: a₂ + a₇ = 30

Also, a₁₅ = 2a₈ - 1

Consider a₂ + a₇ = 30

⇒ (a + d) + (a + 6d) = 30

⇒ 2a + 7d = 30 ……………….. (1)

Consider a₁₅ = 2a₈ - 1

⇒ a + 14d = 2(a + 7d) - 1

⇒ a + 14d = 2a + 14d - 1

⇒ a = 1

∴ First term = a = 1

Thus, from equation (1), we get,

7d = 30 - 2a

⇒ 7d = 30 - 2

⇒ 7d = 28

⇒ d = 4

Thus, the AP is a, a + d, a + 2d, a + 3d,…

Therefore, the AP is 1, 5, 9, 13, 17,…

Let a₁ and d₁ be the first term and common difference of the AP 63, 65, 67, 69,….

Let a₂ and d₂ be the first term and common difference of the AP 3, 10, 17,….

∴ a₁ = 63, d₁ = 2

a₂ = 3, d₂ = 7

Let a_n be the n^th term of the first AP and b_n be the n^th term of the second AP.

So, a_n = a₁ + (n - 1)d₁

⇒ a_n = 63 + (n - 1)2

⇒ a_n = 61 + 2n

and, b_n = a₂ + (n - 1)d₂

⇒ b_n = 3 + (n - 1)7

⇒ b_n = - 4 + 7n

Since for nth terms of both the AP’s to be same, a_n = b_n

⇒ 61 + 2n = - 4 + 7n

⇒ 61 + 4= 7n - 2n

⇒ 65 = 5n

⇒ n = 13

Therefore, 13^th term of both the AP’s will be same.

Let a and d be the first term and common difference of the AP Given: a₁₇ = 2 × a₈ + 5

a₁₁ = 43

To find: n^th term = a_n

Consider a11 = 43

⇒ a + (11 - 1)d = 43

⇒ a + 10d = 43 ………….. (1)

Consider a₁₇ = 2 × a₈ + 5

⇒ a + (17 - 1)d = 2[a + (8 - 1)d] + 5

⇒ a + 16d = 2a + 14d + 5

⇒ - a + 2d = 5 ……………….(2)

Adding equation (1) and equation (2), we get

12d = 48

⇒ d = 4

∴ from equation (1), we get,

a = 43 - 10d

= 43 - 40

= 3

Now, n^th term is given by:

a_n = a + (n - 1)d

⇒ a_n = 3 + (n - 1)4

⇒ a_n = 4n - 1

Therefore, n^th term is given by (4n - 1).

Let a be the first term and d be the common difference.

Given: a₂₄ = 2(a₁₀)

To prove: a₇₂ = 4 × a₁₅

Now, Consider a₂₄ = 2a₁₀

⇒ a + 23d = 2[a + 9d]

⇒ a + 23d = 2a + 18d

⇒ a = 5d …………………. (1)

Consider a₇₂ = a + (72 - 1)d

⇒ a₇₂ = 5d + 71d (from equation (1))

⇒ a₇₂ = 76d ………………. (2)

Now, consider a₁₅ = a + (15 - 1)d

⇒ a₁₅ =5d + 14d (from equation (1))

⇒ a₁₈ = 19d ………………….(3)

From equation (2) and (3), we get,

a₇₂ = 4 × a₁₅

Hence, proved.

Let a be the first term and d be the common difference.

Given: a₉ = 19

a₁₉ = 3 a₆

Now, Consider a₉ = 19

⇒ a + (9 - 1)d = 19

⇒ a + 8d = 19 ………………….(1)

Consider a₁₉ = 3 a₆

⇒ a + 18d = 3(a + 5d)

⇒ a + 18d = 3a + 15d

⇒ 2a - 3d = 0 ………………….(2)

Now, subtracting twice of equation (1) from (2), we get,

- 19d = - 38

⇒ d = 2

∴ from equation (1), we get,

a = 19 - 8d

⇒ a = 19 - 8 × 2

⇒ a = 19 - 16

⇒ a = 3

Thus the AP is a, a + d, a + 2d, a + 3d, a + 4d,….

Therefore the AP is 3, 5, 7, 9….

Let a be the first term and d be common difference.

Given: a_p = q

a_q = p

To show: a_{(p + q)} = 0

We know, nth term of an AP is
a_n = a + (n - 1)d
where, a is first term and d is common difference
Consider a_p = q

⇒ a + (p - 1)d = q (1)

Consider a_q = p

⇒ a + (q - 1)d = p (2)

Now, subtracting equation (2) from equation (1), we get

(p - q)d = (q - p)

⇒ d = - 1

∴ From equation (1), we get,

a - p + 1 = q

⇒ p + q = a + 1 ……………………….(3)

Consider a_{(p + q)} = a + (p + q - 1)d

= a + (p + q - 1)(-1)

= a + (a + 1 - 1)(-1)

(putting the value of p + q from equation 3)

= a + (-a)

= 0

∴ a_{(p + q)} = 0

Hence, proved.

Let d be the common difference of the AP.

First term = a

Last term = l = 1

n^th term from beginning of an AP is given by:

a_n = a + (n - 1)d …………………………….(1)

n^th term from the end of an AP is given by:

T_n = l - (n - 1)d

= 1 - (n - 1)d …………………………….(2)

Sum of the n^th term from the beginning and end is given by:

a_n + T_n = a + (n - 1)d + 1 - (n - 1)d

= a + 1

Hence, proved.

The two digit numbers divisible by 6 are 12, 18, 24, 30,…96.

This forms an AP with first term a = 12

and common difference = d = 6

Last term is 96.

Now, number of terms in this AP are given as:

96 = a + (n - 1)d

⇒ 96 = 12 + (n - 1)6

⇒ 96 - 12 = 6n - 6

⇒ 84 + 6 = 6n

⇒ 90 = 6n

⇒ n = 15

There are 15 two - digit numbers that are divisible by 6.

The two digit numbers divisible by 3 are 12, 15, 18, 21, …., 99.

This forms an AP with first term a = 12

and common difference = d = 3

Last term is 99.

Now, number of terms in this AP are given as:

99 = a + (n - 1)d

⇒ 99 = 12 + (n - 1)3

⇒ 99 - 12 = 3n - 3

⇒ 87 + 3 = 3n

⇒ 90 = 3n

⇒ n = 30

There are 30 two - digit numbers that are divisible by 3.

The three digit numbers divisible by 9 are 108, 117, 126, …., 999.

This forms an AP with first term a = 108

and common difference = d = 9

Last term is 999.

Now, number of terms in this AP are given as:

999 = a + (n - 1)d

⇒ 999 = 108 + (n - 1)9

⇒ 999 - 108 = 9n - 9

⇒ 891 + 9 = 9n

⇒ 900 = 9n

⇒ n = 100

There are 100 three - digit numbers that are divisible by 9.

The numbers between 101 and 999 that are divisible by both 2 and 5 are 110, 120, 130,…, 990.

This forms an AP with first term a = 110

and common difference = d = 10

Last term is 990.

Now, number of terms in this AP are given as:

990 = a + (n - 1)d

⇒ 990 = 110 + (n - 1)10

⇒ 990 - 110 = 10n - 10

⇒ 880 + 10 = 10n

⇒ 890 = 10n

⇒ n = 89

There are 89 numbers between 101 and 999 that are divisible by both 2 and 5.

The no of rose plants in each row can be arranged in the form of an AP as 43, 41, 39, …, 11.

Here, First term = a = 43

Common difference = d = 41 - 43 = - 2

No of terms in the AP = No of rows in the flower bed.

∴ 11 = a + (n - 1)d

⇒ 11 = 43 + (n - 1)(-2)

⇒ 11 - 43 = - 2n + 2

⇒ 11 - 43 - 2 = - 2n

⇒ 2n = 34

⇒ n = 17

∴ No of rows in the flower bed = 17

Let the first prize be Rs. x. Thus each succeeding prize is Rs. 200 less than the preceding prize.

∴ Second prize is Rs. (x - 200)

Third prize is Rs. (x - 400)

Fourth prize is Rs. (x - 600)

This forms an AP as x, x - 200, x - 400, x - 600.

Since, Total sum of prize amount = 2800.

∴ x + (x - 200) + (x - 400) + (x - 600) = 2800

⇒ 4x - 1200 = 2800

⇒ 4x = 2800 + 1200

⇒ 4x = 4000

⇒ x = 1000

Thus, the first, second, third and fourth prizes are as Rs. 1000, Rs. 800, Rs. 600, Rs. 400.

If three terms are in AP, the difference between the terms should be equal, i.e. if a, b and c are in AP then,
b - a = c - b

Since, the terms are in an AP, therefore

(4k - 6) - (3k - 2) = (k + 2) - (4k - 6)

⇒ k - 4 = - 3k + 8

⇒ 4k = 12

⇒ k = 3

∴ k = 3

Given: The numbers (5x + 2), (4x - 1) and (x + 2) are in AP.

To find: The value of x.

Solution:

Let a₁ = (5x + 2)

a₂ = 4x - 1)

a₃ = (x + 2)
Since, the terms are in an AP, therefore common difference is same.

⇒ a₂ - a_{1 =} a₃ - a₂

⇒(4x - 1) - (5x + 2) = (x + 2) - (4x - 1)

⇒ - x - 3 = - 3x + 3

⇒ 2x = 6

⇒ x = 3

∴ x = 3

Since, the terms are in an AP, therefore

(3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)

⇒ 6 = 2y - 4

⇒ 2y = 10

⇒ y = 5

∴ y = 5

Given: (x + 2), 2x, (2x + 3) are three consecutive terms of an AP.

To find: the value of x

Solution:

Let a₁ = x + 2

a₂ = 2x

a₃ = 2x + 3

As, a₁, a₂ and a₃ are in AP, common difference will be equal

⇒ a₂ – a₁ = a₃ – a₂

⇒ (2x) - (x + 2) = (2x + 3) - (2x)

⇒ 2x - x - 2= 2x + 3 - 2x

⇒ x - 2 = 3

⇒ x = 5

Consider (a² + b²) - (a - b)²

= (a² + b²) - (a² + b² - 2ab)

= 2ab

Consider (a + b)² - (a² + b²)

= (a² + b² + 2ab) - (a² + b²)

= 2ab

Since, the difference between consecutive terms is constant, therefore the terms are in AP.

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 15

∴ (a - d) + a + (a + d) = 15

⇒ 3a = 15

⇒ a = 5

Now, product of the numbers = 80

⇒ (a - d) × a × (a + d) = 80

⇒ a³ - ad² = 80

Put the value of a, we get,

125 - 5 d² = 80

⇒ 5 d² = 125 - 80 = 45

d² = 9

d = � 3

∴ If d = 3, then the numbers are 2, 5, 8.

If d = - 3, then the numbers are 8, 5, 2.

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 15

∴ (a - d) + a + (a + d) = 3

⇒ 3a = 3

⇒ a = 1

Now, product of the numbers = - 35

⇒ (a - d) × a × (a + d) = - 35

⇒ a³ - ad² = - 35

Put the value of a, we get,

1 - d² = - 35

⇒ d² = 35 + 1 = 36

d² = 36

d = 6

∴ If d = 6, then the numbers are - 5, 1, 7.

If d = - 6, then the numbers are 7, 1, - 5.

Let 24 be divided in numbers which are in AP as (a - d), a, (a + d).

Now, sum of the numbers = 24

∴ (a - d) + a + (a + d) = 24

⇒ 3a = 24

⇒ a = 8

Now, product of the numbers = 440

⇒ (a - d) × a × (a + d) = 440

⇒ a³ - ad² = 440

Put the value of a, we get,

512 - 8d² = 440

⇒ 8d² = 512 - 440 = 72

d² = 9

d = � 3

∴ If d = 3, then the numbers are 5, 8, 11.

If d = - 3, then the numbers are 11, 8, 5.

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 21

∴ (a - d) + a + (a + d) = 21

⇒ 3a = 21

⇒ a = 7

Now, sum of the squares of the terms = 165

⇒ (a - d)² + a² + (a + d)² = 165

⇒ a² + d² - 2ad + a² + a² + d² + 2ad = 165

⇒ 3a² + 2d² + a = 165

Put the value of a = 7, we get,

3(49) + 2d² = 165

⇒ 2d² = 165 - 147

⇒ 2d² = 18

⇒ d² = 9

⇒ d = 3

∴ If d = 3, then the numbers are 4, 7, 10.

If d = - 3, then the numbers are 10, 7, 4.

Let these angles be x°, (x + 10)°, (x + 20)° and (x + 30)°.

Since, Sum of all angles of a quadrilateral = 360°.

⇒ x° + (x + 10)° + (x + 20)° + (x + 30)° = 360°

⇒ 4x + 60° = 360°

⇒ 4x = 300°

⇒ x = 75°

∴ the angles will be 75°, 85°, 95°, 105°.

Let the numbers be (a - 3d), (a - d), (a + d), (a + 3d).

Now, sum of the numbers = 28

∴ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 28

⇒ 4a = 28

⇒ a = 7

Now, sum of the squares of the terms = 216

⇒ (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)²= 216

⇒ a² + 9d² - 6ad + a² + d² - 2ad + a² + d² + 2ad + a² + 9d² + 6ad = 216

⇒ 4a² + 20d² = 216

Put the value of a = 7, we get,

4(49) + 20d² = 216

⇒ 20d² = 216 - 196

⇒ 20d² = 20

⇒ d² = 1

⇒ d = 1

∴ If d = 1, then the numbers are 4, 6, 8, 10.

If d = - 1, then the numbers are 10, 8, 6, 4.

Let 32 be divided into parts as (a - 3d), (a - d), (a + d) and (a + 3d).

Now (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32

⇒ a = 8

Now, we are given that product of the first and the fourth terms is to the product of the second and the third terms as 7 : 15.

i.e. [(a - 3d) × (a + 3 d)] : [(a - d) × (a + d)] = 7 : 15

⇒ =

⇒ 15[(a - 3d) × (a + 3 d)] = 7[(a - d) × (a + d)]

⇒ 15[a² - 9d²] = 7 [a² - d²]

⇒ 15a² - 135d² = 7a² - 7d²

⇒ 8a² - 128d² = 0

⇒ 8a² = 128d²

Putting the value of a, we get,

512 = 128 d²

⇒ d² = 4

⇒ d = ±2

∴ If d = 2, then the numbers are 2, 6, 10, 14.

If d = - 2, then the numbers are 14, 10, 6, 2.

Let the numbers be (a - d), a, (a + d).

Now, sum of the numbers = 48

∴ (a - d) + a + (a + d) = 48

⇒ 3a = 48

⇒ a = 16

Now, we are given that,

Product of first and second terms exceeds 4 times the third term by 12.

⇒ (a - d) × a = 4(a + d) + 12

⇒ a² - ad = 4a + 4d + 12

On putting the value of a in the above equation, we get,

256 - 16d = 64 + 4d + 12

⇒ 20 d = 180

⇒ d = 9

∴ The numbers are a - d, a, a + d

i.e. the numbers are 7, 16, 25.

Since, the terms are in an AP, therefore

(3y + 5) - (3y - 1) = (5y + 1) - (3y + 5)

⇒ 6 = 2y - 4

⇒ 2y = 10

⇒ y = 5

∴ y = 5

Since, the terms are in an AP, therefore

(2k - 1) - k = (2k + 1) - (2k - 1)

⇒ k - 1 = 2

⇒ k = 3

∴ k = 3

Since, the terms are in an AP, therefore

a - 18 = (b - 3) - a

⇒ 2a - b = - 3 + 18

⇒ 2a - b = 15

∴ 2a - b = 15

Since, the terms are in an AP, therefore

9 - a = b - 9 = 25 - b

Consider b - 9 = 25 - b

⇒ 2b = 34

⇒ b = 17

Now, consider the first equality,

9 - a = b - 9

⇒ a = 18 - b

⇒ a = 18 - 17

⇒ a = 1

∴ a = 1, b = 17

Since, the terms are in an AP, therefore

(3n + 2) - (2n - 1) = (6n - 1) - (3n + 2)

⇒ n + 3 = 3n - 3

⇒ 2n = 6

⇒ n = 3

∴ n= 3, and hence the numbers are 5, 11, 17.

The three digit numbers divisible by 7 are 105, 112, 119, …., 994.

This forms an AP with first term a = 105

and common difference = d = 7

Last term is 994.

Now, number of terms in this AP are given as:

994 = a + (n - 1)d

⇒ 994 = 105 + (n - 1)7

⇒ 994 - 105 = 7n - 7

⇒ 889 + 7 = 7n

⇒ 896 = 7n

⇒ n = 128

Therefore 994 is the 128^th term in the AP.

∴ There are 128 three - digit natural numbers that are divisible by 7.

The three digit natural numbers divisible by 9 are 108, 117, 126, …., 999.

This forms an AP with first term a = 108

and common difference = d = 9

Last term is 999.

Now, number of terms in this AP are given as:

999 = a + (n - 1)d

⇒ 999 = 108 + (n - 1)9

⇒ 999 - 108 = 9n - 9

⇒ 891 + 9 = 9n

⇒ 900 = 9n

⇒ n = 100

Therefore 999 is the 100^th term in the AP.

∴ There are 100 three - digit natural numbers that are divisible by 9.

Let S_n denotes the sum of first n terms of an AP.

Sum of first m terms = S_m = 2m² + 3m

Then n^th term is given by: a_n = S_n - S_{n - 1}

We need to find the 2^nd term, so put n = 2, we get

a₂ = S₂ - S₁

= (2(2)² + 3(2)) - (2(1)² + 3(1))

= 14 - 5

= 9

∴ the second term of the AP is 9.

Here, first term = a

Common difference = 3a - a = 2a

Now, Sum of first n terms of an AP is given by:

S_n = [2a + (n - 1)d]

where a is the first term and d is the common difference.

∴ Sum of first n terms of given AP is given by:

S_n = [2a + (n - 1)2a]

= [2a + 2an - 2a]

= [2an]

= n²a

Here, First term = a = 2

Common difference = d = 7 - 2 = 5

Last term = l = 47

To find:5^th term from end.

So, n^th term from end is given by:

a_n = l - (n - 1)d

∴ 5^th term from end is:

a₅ = 47 - (5 - 1) × 5

= 47 - 20

= 27

∴ 5^th term from the end is 27.

Here, First term = a = 2

Common difference = d = 7 - 2 = 5

To find: a₃₀ – a₂₀

So, n^th term is given by:

a_n = a + (n - 1)d

∴ 30^th term is:

a₃₀ = 2 + (30 - 1) × 5

= 2 + 145

= 147

Now, 20^th term is:

a₂₀ = 2 + (20 - 1) × 5

= 2 + 95

= 97

Now, (a₃₀ – a₂₀) = 147 - 97

= 50

∴ (a₃₀ – a₂₀) = 50

n^th term of an AP = a_n = 3n + 5

Common difference (= d) of an AP is the difference between a term and its preceding term.

∴ d = a_n - a_{n - 1}

= (3n + 5) - (3(n - 1) + 5)

= 3n + 5 - 3n + 3 - 5

= 3

∴ Common difference = 3

n^th term of an AP = a_n = 7 - 4n

Common difference (= d) of an AP is the difference between a term and its preceding term.

∴ d = a_n - a_{n - 1}

= (7 - 4n) - (7 - 4(n - 1))

= 7 - 4n - 7 + 4n - 4

= - 4

∴ Common difference = - 4.

Here, first term = √8

Common difference = √18 - √8 = √2

Next term = T₄ = T₃ + d

= √32 + √2

= 4√2 + √2

= 5√2

= √50

Here, first term = √2

Common difference = √8 - √2 = 2√2 - √2 = √2

Next term = T₄ = T₃ + d

= √18 + √2

= 3√2 + √2

= 4√2

= √32

Here first term = 21

Common difference = 18 - 21 = - 3

Let a_n be the term which is zero.

∴ a_n = 0

⇒ a + (n - 1)d = 0

⇒ 21 + (n - 1)(-3) = 0

⇒ 21 - 3n + 3 = 0

⇒ 3n = 24

⇒ n = 8

∴ 8^th term of the given AP will be zero.

First n natural numbers are 1, 2, 3,…, n.

To find: sum of these n natural numbers.

The natural numbers forms an AP with first term 1 and common difference 1.

Now, Sum of first n terms of an AP is given by:

S_n = [2a + (n - 1)d]

where a is the first term and d is the common difference.

∴ Sum of first n natural numbers is given by:

S_n = [2(1) + (n - 1)(1)]

= [2 + n - 1]

= [n + 1]

∴ Sum of first n natural numbers is n(n + 1)/2.

First n even natural numbers are 2, 4, 6,…, 2n.

To find: sum of these n even natural numbers.

The even natural numbers forms an AP with first term 2 and common difference 2.

Now, Sum of first n terms of an AP is given by:

S_n = [2a + (n - 1)d]

where a is the first term and d is the common difference.

∴ Sum of first n natural numbers is given by:

S_n = [2(2) + (n - 1)(2)]

= [4 + 2n - 2]

= [2n + 2]

= n (n + 1)

∴ Sum of first n even natural numbers is n(n + 1).

Here, given: first term = p

Common difference = q

To find: a₁₀

a₁₀ = a + (10 - 1)d

⇒ a₁₀ = p + 9q

∴ 10^th term of the given AP will be p + 9q.

Since, the terms are in an AP, therefore

a - (4/5) = 2 - a

⇒ 2a = 2 + (4/5)

⇒ 2a = 14/5

⇒ a = 14/10

⇒ a = 7/5

∴ a = 7/5

Since, the terms are in an AP, therefore

13 - (2p + 1) = (5p - 3) - (13)

⇒ 12 - 2p = 5p - 16

⇒ 7p = 28

⇒ p = 4

∴ p = 4

Since, the terms are in an AP, therefore

7 - (2p - 1) = 3p - 7

⇒ 8 - 2p = 3p - 7

⇒ 5p = 15

⇒ p = 3

∴ p = 3

Let S_p denotes the sum of first p terms of an AP.

Sum of first p terms = S_p = ap² + bp

Then p^th term is given by: a_p = S_p - S_{p - 1}

∴ a_p = (ap² + bp) - [a(p - 1)² + b(p - 1)]

= (ap² + bp) - [a(p² + 1 - 2p) + bp - b]

= ap² + bp - ap² - a + 2ap - bp + b

= b - a + 2ap

Now, common difference = d = a_p - a_{p - 1}

= b - a + 2ap - [b - a + 2a(p - 1)]

= b - a + 2ap - b + a - 2ap + 2a

= 2a

∴ common difference = 2a

ALITER: Let S_p denotes the sum of first p terms of an AP.

Sum of first p terms = S_p = ap² + bp

Put p = 1, we get S₁ = a + b

Put p = 2, we get S₂ = 4a + 2b

Now S₁ = a₁

a₂ = S₂ - S₁

∴ a₂ = 3a + b

Now, d = a₂ - a₁

= 3a + b - (a + b)

= 2a

∴ Common difference = 2a

Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 3n² + 5n

Then n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (3n² + 5n) - [3(n - 1)² + 5(n - 1)]

= (3n² + 5n) - [3(n² + 1 - 2n) + 5n - 5]

= 3n² + 5n - 3n² - 3 + 6n - 5n + 5

= 2 + 6n

Now, common difference = d = a_n - a_{n - 1}

= 2 + 6n - [2 + 6(n - 1)]

= 2 + 6n - 2 - 6n + 6

= 6

∴ Common difference = 6

ALITER: Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 3n² + 5n

Put n = 1, we get S₁ = 8

Put n = 2, we get S₂ = 22

Now S₁ = a₁

a₂ = S₂ - S₁

∴ a₂ = 22 - 8 = 14

Now, d = a₂ - a₁

= 14 - 8

= 6

∴ Common difference = 6

Let a be the first term and d be the common difference.

Given: a₄ = 9

a₆ + a₁₃ = 40

Now, Consider a₄ = 9

⇒ a + (4 - 1)d = 9

⇒ a + 3d = 9 ……….(1)

Consider a₆ + a₁₃ = 40

⇒ a + (6 - 1)d + a + (13 - 1)d = 40

⇒ 2a + 17d = 40 ………..(2)

Subtracting twice of equation (1) from equation (2), we get,

11d = 22

⇒ d = 2

∴ Common difference = d = 2

Now from equation (1),we get

a = 9 - 3d

= 9 - 6

= 3

∴ AP is a, a + d, a + 2d, a + 3d, …

i.e. AP is 3, 5, 7,9, 11…..

Here, first term = 2

Common difference = 7 - 2 = 5

Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ S₁₉ = [2(2) + (19 - 1)5]

= (19)(4 + 90)/2

= (19 × 94)/2

= 893

Thus, sum of 19 terms of this AP is 893.

Here, first term = 9

Common difference = 7 - 9 = - 2

Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ S₁₄ = [2(9) + (14 - 1)(-2)]

= (7)(18 - 26)

= (7) × (-8)

= - 56

Thus, sum of 14 terms of this AP is - 56.

Here, first term = - 37

Common difference = (-33) - (-37) = 4

Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ S₁₂ = [2(-37) + (12 - 1)(4)]

= (6)(-74 + 44)

= 6 × (-30)

= - 180

Thus, sum of 12 terms of this AP is - 180.

Here, first term = 1/15

Common difference = (1/12) - (1/15) = 1/60

Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ S₁₁ = [2(1/15) + (11 - 1)(1/60)]

= (11/2) × [(2/15) + (1/6)]

= (11/2) × [(3/10)]

= 33/20

Thus, sum of 11 terms of this AP is 33/20.

Here, first term = 0.6

Common difference = 1.7 - 0.6 = 1.1

Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ S₁₀₀ = [2(0.6) + (100 - 1)(1.1)]

= (50) × [1.2 + (99 × 1.1) ]

= 50 × [1.2 + 108.9]

= 50 × 110.1

= 5505

Thus, sum of 100 terms of this AP is 5505.

Here, First term = 7

Common difference = d = (21/2) - 7 = (7/2)

Last term = l = 84

Now, 84 = a + (n - 1)d

∴ 84 = 7 + (n - 1)(7/2)

⇒ 84 - 7 = (n - 1)(7/2)

⇒ 77 = (n - 1)(7/2)

⇒ 154 = 7n - 7 (multiplying both sides by 2)

⇒ 154 + 7 = 7n

⇒ 7n = 161

⇒ n = 23

∴ there are 23 terms in this Arithmetic series.

Now, Sum of these 23 terms is given by

∴ S₂₃ = [2(7) + (23 - 1)(7/2)]

= (23/2) × [14 + (22)(7/2) ]

= (23/2) × [14 + 77]

= (23/2) × [91]

= 2093/2

= 1046.5

Thus, sum of 23 terms of this AP is 1046.5.

Here, First term = 34

Common difference = d = 34 - 32 = - 2

Last term = l = 10

Now, 10 = a + (n - 1)d

∴ 10 = 34 + (n - 1)(-2)

⇒ 10 - 34 = (n - 1)(-2)

⇒ - 24 = - 2n + 2

⇒ - 24 - 2 = - 2n

⇒ - 26 = - 2n

⇒ n = 13

⇒ n = 13

∴ there are 13 terms in this Arithmetic series.

Now, Sum of these 13 terms is given by

∴ S₁₃ = [2(34) + (13 - 1)(-2)]

= (13/2) × [68 + (12)(-2) ]

= (13/2) × [68 - 24]

= (13/2) × [44]

= 13 × 22

= 286

Thus, sum of 23 terms of this AP is 286.

Here, First term = - 5

Common difference = d = - 8 - (-5) = - 3

Last term = l = - 230

Now, - 230 = a + (n - 1)d

∴ - 230 = - 5 + (n - 1)(-3)

⇒ - 230 + 5 = (n - 1)(-3)

⇒ - 225 = - 3n + 3

⇒ - 225 - 3 = - 3n

⇒ - 228 = - 3n

⇒ n = 76

∴ there are 76 terms in this Arithmetic series.

Now, Sum of these 76 terms is given by

∴ S₇₆ = [2(-5) + (76 - 1)(-3)]

= 38 × [ - 10 + (75)(-3) ]

= 38 × [ - 10 - 225]

= 38 × (-235)

= - 8930

Thus, sum of 23 terms of this AP - 8930.

Since, nth term is given as (5 - 6n)

Put n = 1, we get a₁ = - 1 = first term

Put n = 2, we get a₂ = - 7 = second term

Now, d = a₂ - a₁ = - 7 - (-1) = - 6

Sum of first n terms = S_n= [2a + (n - 1)d]; where a is the first term

and d is the common difference.

= [ - 2 + (n - 1)(-6)]

= n[ - 1 - 3n + 3]

= n(2 - 3n)

∴ sum of first 20 terms = S₂₀

= [2(-1) + (20 - 1)(-6)]

= 10 × [ - 2 - 114]

= 10 × [ - 116]

= - 1160

Given: The sum of the first n terms of an AP is (3n² + 6n).

To find: the nth term and the 15th term of this AP.

Solution:

Sum of first n terms = S_n = 3n² + 6n

Now let a_n be the n^th term of the AP.

To find: a_n and a₁₅

Since a_n =S_n - S_{n - 1}


⇒ a_n= (3n² + 6n) - (3(n - 1)² + 6(n - 1))

⇒ a_n= (3n² + 6n) - (3(n² +1 - 2n) + 6(n - 1))

⇒ a_n = (3n² + 6n) - (3n² + 3 - 6n + 6n - 6)

⇒ a_n = 6n + 3

Now, a₁₅ = 6(15) + 3

⇒ a₁₅= 93

(i) Let a_n be the n^th term of the AP.

To find: a_n

Then a_n =S_n - S_{n - 1}

= (3n² - n) - (3(n - 1)² - (n - 1))

= (3n² - n) - (3n² + 3 - 6n - n + 1)

= 6n - 4

(ii) Since a_n = 6n - 4

∴ For first term, n = 1

By putting n = 1 in the nt^h term, we get,

a₁ = 6(1) - 4

= 2

∴ a = 2

(iii) Put n = 2 in the nth term, we get

a₂ = 6 × (2) - 4

= 12 - 4

= 8

Now common difference = d = a₂ - a₁

= 8 - 2

= 6

∴ Common difference = 6

Let a_n be the n^th term of the AP.

To find: a_n and a₂₀

Since, a_n =S_n - S_{n - 1}

= ( ) - ()

= 1/2 (5n² + 3n) - 1/2 [5(n - 1)² + 3(n - 1)]

= 1/2 (5n² + 3n - 5n² - 5 + 10n - 3n + 3)

= 1/2 (10n - 2)

= 5n - 1

Since a_n = 5n - 1

∴ For 20^th term, put n = 20, we get,

a₂₀ = 5(20) - 1

= 100 - 1

= 99

Let a_n be the n^th term of the AP.

To find: a_n and a₂₅

Since, a_n =S_n - S_{n - 1}

= ( ) - ()

= 1/2 (3n² + 5n) - 1/2 [3(n - 1)² + 5(n - 1)]

= 1/2 (3n² + 5n - 3n² - 3 + 6n - 5n + 5)

= 1/2 (6n - 2)

= 3n + 1

Since a_n = 5n - 1

∴ For 25^th term, put n = 25, we get,

a₂₅ = 3(25) + 1

= 75 + 1

= 76

Here, first term = a = 21

Common difference = d = 18 - 21 = - 3

Let first n terms of the AP sums to zero.

∴ S_n = 0

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 0

∴ (n/2) × [2a + (n - 1)d] = 0

⇒ (n/2) × [2(21) + (n - 1)(-3)] = 0

⇒ (n/2) × [42 - 3n + 3)] = 0

⇒ (n/2) × [45 - 3n] = 0

⇒ [45 - 3n] = 0

⇒ 45 = 3n

⇒ n = 15

∴ 15 terms of the given AP sums to zero.

Here, first term = a = 9

Common difference = d = 17 - 9 = 8

Let first n terms of the AP sums to 636.

∴ S_n = 636

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 636

∴ (n/2) × [2a + (n - 1)d] = 636

⇒ (n/2) × [2(9) + (n - 1)(8)] = 636

⇒ (n/2) × [18 + 8n - 8)] = 636

⇒ (n/2) × [10 + 8n] = 636

⇒ n[5 + 4n] = 636

⇒ 4n² + 5n - 636 = 0

⇒ 4n² + 5n - 636 = 0

⇒ (n - 12)(4n + 53) = 0

⇒ n = 12 or n = - 53/4

But n can’t be negative and fraction.

∴ n= 12

∴ 12 terms of the given AP sums to 636.

Here, first term = a =63

Common difference = d = 60 - 63 = - 3

Let first n terms of the AP sums to 693.

∴ S_n = 693

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 693

∴ (n/2) × [2a + (n - 1)d] = 693

⇒ (n/2) × [2(63) + (n - 1)(-3)] = 693

⇒ (n/2) × [126 - 3n + 3)] = 693

⇒ (n/2) × [129 - 3n] = 693

⇒ n[129 - 3n] = 1386

⇒ 129n - 3n² = 1386

⇒ 3n² - 129n + 1386 = 0

⇒ (n - 22)( n - 21)= 0

⇒ n = 22 or n = 21

∴ n= 22 or n = 21

Since, a₂₂ = a + 21d

= 63 + 21(-3)

= 0

∴ Both the first 21 terms and 22 terms give the sum 693 because the 22^nd term is 0. So, the sum doesn’t get affected.

Here, first term = a =20

Common difference = d = 58/3 - 20 = - 2/3

Let first n terms of the AP sums to 300.

∴ S_n = 300

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 300

∴ (n/2) × [2a + (n - 1)d] = 300

⇒ (n/2) × [2(20) + (n - 1)(-2/3)] = 300

⇒ (n/2) × [40 - (2/3)n + (2/3)] = 300

⇒ (n/2) × [(120 - 2n + 2)/3] = 300

⇒ n[122 - 2n] = 1800

⇒ 122n - 2n² = 1800

⇒ 2n² - 122n + 1800 = 0

⇒ n² - 61n + 900 = 0

⇒ (n - 36)( n - 25)= 0

⇒ n = 36 or n = 25

∴ n= 36 or n = 25

Now, S₃₆ = (36/2)[2a + 35d]

= 18(40 + 35(-2/3))

= 18(120 - 70)/3

= 6(50)

= 300

Also, S₂₅ = (25/2)[2a + 24d]

= (25/2)(40 + 24(-2/3))

= (25/2)(40 - 16)

= (24 × 25)/2

= 12 × 25

= 300

Now, sum of 11 terms from 26^th term to 36^th term = S₃₆ - S₂₅ = 0

∴ Both the first 25 terms and 36 terms give the sum 300 because the sum of last 11 terms is 0. So, the sum doesn’t get affected.

Odd numbers from 0 to 50 are 1, 3, 5, …, 49

Sum of these numbers is 1 + 3 + 5 + … + 49.

This forms an Arithmetic Series with first term = a = 1

and Common Difference = d = 3 - 1 = 2

There are 25 terms in this Arithmetic Series.

Now, sum of n terms is given as:

S_n = (n/2)[2a + (n - 1)d]

S₂₅ = (25/2)[2(1) + (25 - 1)2]

= (25/2)[2 + 48]

= (25 × 50)/2

= 25 × 25

= 625

∴ Sum of odd numbers from 0 to 50 is 625.

Natural numbers between 200 and 400 which are divisible by 7 are 203, 210, 217, …, 399.

Sum of these numbers forms an arithmetic series 203 + 210 + 217 + … + 399.

Here, first term = a = 203

Common difference = d = 7

∴ a_n = a + (n - 1)d

⇒ 399 = 203 + (n - 1)7

⇒ 399 = 7n + 196

⇒ 7n = 203

⇒ n = 29

∴ there are 29 terms in the AP.

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 28 terms of this arithmetic series is given by:

∴ S₂₉ = [2(203) + (29 - 1)(7)]

= (29/2) [406 + 196]

=(29/2) × 502

= 7279

First 40 positive integers divisible by 6 are 6, 12, 18, …, 240.

Sum of these numbers forms an arithmetic series 6 + 12 + 18 + … + 240.

Here, first term = a = 6

Common difference = d = 6

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 40 terms of this arithmetic series is given by:

∴ S₄₀ = [2(6) + (40 - 1)(6)]

= 20 [12 + 234]

=20 × 246

= 4920

First 15 multiples of 8 are 8, 16, 24, …, 120.

Sum of these numbers forms an arithmetic series 8 + 16 + 24 + … + 120.

Here, first term = a = 8

Common difference = d = 8

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 15 terms of this arithmetic series is given by:

∴ S₁₅ = [2(8) + (15 - 1)(8)]

= (15/2) [16 + 112]

=(15/2) × 128

= 15 × 64

= 960

Multiples of 9 lying between 300 and 700 are 306, 315, 324, …, 693.

Sum of these numbers forms an arithmetic series 306 + 315 + 324 + … + 693.

Here, first term = a = 306

Common difference = d = 9

We first find the number of terms in the series.

Here, last term = l = 693

∴ 693 = a + (n - 1)d

⇒ 693 = 306 + (n - 1)9

⇒ 693 - 306 = 9n - 9

⇒ 387 = 9n - 9

⇒ 387 + 9 = 9n

⇒ 9n = 396

⇒ n = 44

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 44 terms of this arithmetic series is given by:

∴ S₄₄ = [2(306) + (44 - 1)(9)]

= 22 × [612 + 387]

= 22 × 999

= 21978

Three - digit natural numbers which are divisible by 13 are 104, 117, 130, …, 988.

Sum of these numbers forms an arithmetic series 104 + 117 + 130 + … + 988.

Here, first term = a = 104

Common difference = d = 13

We first find the number of terms in the series.

Here, last term = l = 988

∴ 988 = a + (n - 1)d

⇒ 988 = 104 + (n - 1)13

⇒ 988 - 104 = 13n - 13

⇒ 884 = 13n - 13

⇒ 884 + 13 = 13n

⇒ 13n = 897

⇒ n = 69

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 69 terms of this arithmetic series is given by:

∴ S₆₉ = [2(104) + (69 - 1)(13)]

= (69/2) × [208 + 884]

= (69/2) × 1092

= 69 × 546

= 3767

First 100 even natural numbers which are divisible by 5 are 10, 20, 30, …, 1000

Here, first term = a = 10

Common difference = d = 10

Number of terms = 100

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 100 terms of this arithmetic series is given by:

∴ S₁₀₀ = [2(10) + (100 - 1)(10)]

= 50 × [20 + 990]

= 50 × 1010

= 50500

The given sum can be written as (1 + 1 + 1 + …) - (1/n, 2/n, 3/n, …)

Sum of first series up to n terms = 1 + 1 + 1 + … up to n terms

= n

Now, consider the second series:

Here, first term = a = 1/n

Common difference = d = (2/n) - (1/n) = (1/n)

Now, Sum of n terms of an arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of n terms of second arithmetic series is given by:

∴ S_n = [2(1/n) + (n - 1)(1/n)]

= [(2/n) + 1 - (1/n)]

= [(1/n) + 1]

= = = (n + 1)/2

Now, sum of n terms of the complete series = Sum of n terms of first series - Sum of n terms of second series

= n - (n + 1)/2

= (2n - n - 1)/2

= 1/2 (n - 1)

Let the first term be a.

Let Common difference be d.

Given: S₅ + S₇ =167

S₁₀= 235

Now, Sum of n terms of an arithmetic series is given by:

S_n = [2a + (n - 1)d]

So, consider

S₅ + S₇ =167

⇒ (5/2) [2a + (5 - 1)d] + (7/2) [2a + (7 - 1)d] = 167

⇒ (5/2) [2a + 4d] + (7/2) [2a + 6d] = 167

⇒ 5 × [a + 2d] + 7 × [a + 3d] = 167

⇒ 5a + 10d + 7a + 21d = 167

⇒ 12a + 31d = 167 …………….. (1)

Now, consider S₁₀= 235

⇒ (10/2) [2a + (10 - 1)d] = 235

⇒ 5 × [2a + 9d] = 235

⇒ 10a + 45d = 235

⇒ 2a + 9d = 47 ………. (2)

Subtracting equation (1) from 6 times of equation (2), we get,

⇒ 23d = 115

⇒ d = 5

So, from equation (2), we get,

a = 1/2 (47 - 9d)

⇒ a = 1/2 (47 - 45)

⇒ a = 1/2 (2)

⇒ a = 1

Therefore the AP is a, a + d, a + 2d, a + 3d,…

i.e. 1, 6, 11, 16, ….

Here, first term = a = 2

Let the Common difference = d

Last term = l = 29

Sum of all terms = S_n = 155

Let there be n terms in the AP.

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

= [a + a + (n - 1)d]

= [a + l]

Therefore sum of n terms of this arithmetic series is given by:

∴ S_n = [2 + 29] = 155

⇒ 31n = 310

⇒ n = 10

∴ there are 10 terms in the AP.

Thus 29 be the 10^th term of the AP.

∴ 29 = a + (10 - 1)d

⇒ 29 = 2 + 9d

⇒ 27 = 9d

⇒ d = 3

∴ common difference = d =3

Here, first term = a = - 4

Let the Common difference = d

Last term = l = 29

Sum of all terms = S_n = 150

Let there be n terms in the AP.

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

= [a + a + (n - 1)d]

= [a + l]

Therefore sum of n terms of this arithmetic series is given by:

∴ S_n = [ - 4 + 29] = 150

⇒ 25n = 300

⇒ n = 12

∴ there are 12 terms in the AP.

Thus 29 is the 12^th term of the AP.

∴ 29 = a + (12 - 1)d

⇒ 29 = - 4 + 11d

⇒ 29 + 4 = 11d

⇒ 11d = 33

⇒ d = 3

∴ Common difference = d =3

Here, first term = a = 17

Common difference = 9

Last term = l = 350

To find: number of terms and their sum.

Let there be n terms in the AP.

Since, l= 350

∴ 350 = 17 + (n - 1)9

⇒ 350 - 17 = 9n - 9

⇒ 333 = 9n - 9

⇒ 333 + 9 = 9n

⇒ 9n = 342

⇒ n = 38

Therefore number of terms = 38

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

= [a + a + (n - 1)d]

= [a + l]

Therefore sum of 38 terms of this arithmetic series is given by:

∴ S₃₈ = [17 + 350]

= 19 × 367

= 6973

∴ n= 38 and S_n = 6973

Here, first term = a = 5

Let the Common difference = d

Last term = l = 45

Sum of all terms = S_n = 400

Let there be n terms in the AP.

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

= [a + a + (n - 1)d]

= [a + l]

Therefore sum of n terms of this arithmetic series is given by:

∴ S_n = [5 + 45] = 400

⇒ 50n = 800

⇒ n = 16

∴ there are 16 terms in the AP.

Thus 45 is the 16^th term of the AP.

∴ 45 = a + (16 - 1)d

⇒ 45 = 5 + 15d

⇒ 40 = 15d

⇒ 15d = 40

⇒ d = 8/3

∴ Common difference = d = 8/3

Here, first term = a = 22

Let the Common difference = d

n^th term = a_n = - 11

Sum of first n terms = S_n = 66

Let there be n terms in the AP.

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

= [a + a + (n - 1)d]

= [a + a_n]

Therefore sum of n terms of this arithmetic series is given by:

∴ S_n = [22 + (-11)] = 66

⇒ 11n = 132

⇒ n = 12

∴ there are 12 terms in the AP.

Thus n^th is the 12^th term of the AP.

∴ - 11 = a + (12 - 1)d

⇒ - 11 = 22 + 11d

⇒ - 11 - 22 = 11d

⇒ 11d = - 33

⇒ d = - 3

∴ Common difference = d = - 3

∴ n = 12, d = - 3

Let a be the first term and d be the common difference.

Given: a₁₂ = - 13

S₄ = 24

To find: Sum of first 10 terms.

Consider a₁₂ = - 13

⇒ a + 11d = - 13 ………………(1)

Also, S₄ = 24

⇒ (4/2) × [2a + (4 - 1)d] = 24

⇒ 2 × [2a + 3d] = 24

⇒ 2a + 3d = 12 …………….(2)

Subtracting equation (2) from twice of equation (1), we get,

19d = - 38

⇒ d = - 2

Now, from equation (1), we get

a = - 13 - 11d

⇒ a = - 13 - 11(-2)

⇒ a = - 13 + 22

⇒ a = 9

Now, Sum of first n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of first 10 terms of this arithmetic series is given by:

∴ S₁₀ = [2(9) + (10 - 1)(-2)]

= 5 × [18 - 18]

= 0

∴ S₁₀ = 0

Let a be the first term and d be the common difference.

Given: S₇ = 182

4th and 17th terms are in the ratio 1 : 5.

i.e. [a + 3d] : [(a + 16d] = 1 : 5

⇒ =

⇒ 5(a + 3 d) = (a + 16d)

⇒ 5a + 15d = a + 16d

⇒ 4a = d

Now, consider S₇ = 182

⇒ (7/2)[2a + (7 - 1)d] = 182

⇒ (7/2)[2a + 6(4a)] = 182

⇒ 7 × [26a] = 182 × 2

⇒ 182a = 364

⇒ a = 2

∴ d = 4a

⇒ d = 8

Thus the AP will be a, a + d, a + 2d,…

i.e. AP is 2, 10, 18, 26,….

Let a be the first term and d be the common difference.

Given: S₉ = 81, S₂₀ = 400

Now, consider S₉ = 81

⇒ (9/2)[2a + (9 - 1)d] = 81

⇒ (9/2)[2a + 8d] = 81

⇒ [2a + 8d] = 18 ………(1)

Now, consider S₂₀ = 400

⇒ (20/2)[2a + (20 - 1)d] = 400

⇒ 10 × [2a + 19d] = 400

⇒ [2a + 19d] = 40 ………..(2)

Now, on subtracting equation (2) from equation (1), we get,

11d = 22

⇒ d = 2

∴ from equation (1), we get

a = 1/2 (18 - 8d)

⇒ a = 9 - 4d

⇒ a = 9 - 8

⇒ a = 1

∴ a = 1, d = 2

Let a be the first term and d be the common difference.

Given: S₇ = 49, S₁₇ = 289

To find: sum of first n terms.

Now, consider S₇ = 49

⇒ (7/2)[2a + (7 - 1)d] = 49

⇒ (7/2)[2a + 6d] = 49

⇒ [a + 3d] = 7 …………(1)

Now, consider S₁₇ = 289

⇒ (17/2)[2a + (17 - 1)d] = 289

⇒ (17/2) × [2a + 16d] = 289

⇒ [a + 8d] = 17 …………..(2)

Now, on subtracting equation (2) from equation (1), we get,

5d = 10

⇒ d = 2

∴ from equation (1), we get

a = (7 - 3d)

⇒ a = 7 - 6

⇒ a = 1

∴ a = 1, d = 2

Now, Sum of first n terms = S_n = (n/2)[2a + (n - 1)d]

= (n/2)[2 + (n - 1)2]

= (n/2)[2n]

= n²

∴ S_n = n²

Let a₁ and a₂ be the first terms of the two APs

Let d₁ and d₂ be the common difference of the respective APs.

Given: d₁ = d₂ and a₁ = 3, a₂ = 8

To find: Difference between the sums of their first 50 terms.

i.e. to find: (S₂)₅₀ - (S₁)₅₀

where (S₁)₅₀ denotes the sum of first 50 terms of first AP and (S₂)₅₀

denotes the sum of first 50 terms of second AP.

Now, consider (S₁)₅₀ = (50/2)[2a₁ + (50 - 1)d₁]

= 25 × [2(3) + 49 × d₁]

= 25[6 + 49d₁]

= 150 + 1225d₁

Now, consider (S₂)₅₀ = (50/2)[2a₂ + (50 - 1)d₂]

= 25 × [2(8) + 49 × d₂]

= 25[16 + 49d₁]

= 400 + 1225d₂

Now, (S₂)₅₀ - (S₁)₅₀ = 400 + 1225d₂ - (150 + 1225d₂)

= 400 - 150 (∵ d₁ = d₂ )

= 250

∴ (S₂)₅₀ - (S₁)₅₀ = 250

Let a be the first term and d be the common difference.

Given: Sum of first 10 terms = S₁₀ = - 150

Sum of next 10 terms = - 550

i.e. S₂₀ - S₁₀ = - 550

Consider S₁₀ = - 150

⇒ (10/2)[2a + (10 - 1)d] = - 150

⇒ 5 × [2a + 9d] = - 150

⇒ [2a + 9d] = - 30 ………………….(1)

Now, consider S₂₀ - S₁₀ = - 550

⇒ (20/2)[2a + (20 - 1)d] - (10/2)[2a + (10 - 1)d] = - 550

⇒ 10 × [2a + 19d] - 5[2a + 9d] = - 550

⇒ 10a + 145d = - 550 …………………..(2)

On subtracting equation (2) from 5 times of equation (2), we get,

- 100d = 400

⇒ d = - 4

∴ a = 1/2 (-30 - 9d)

⇒ a = 1/2 (-30 + 36)

⇒ a = 3

Therefore the AP is 3, - 1, - 5, - 9,….

Let a be the first term and d be the common difference.

Given: a₅ = 16

a₁₃ = 4 a₃

Now, Consider a₅ = 16

⇒ a + (5 - 1)d = 16

⇒ a + 4d = 16 ………………….(1)

Consider a₁₃ = 4 a₃

⇒ a + 12d = 4(a + 2d)

⇒ a + 12d = 4a + 8d

⇒ 3a - 4d = 0 ………………….(2)

Now, adding equation (1) and (2), we get,

4a = 16

⇒ a = 4

∴ from equation (2), we get,

4d = 3a

⇒ 4d = 12

⇒ d = 3

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 10 terms is given by:

S₁₀ = [2(4) + (10 - 1)(3)]

= 5 × [8 + 27]

= 5 × 35

= 175

∴ S₁₀=175

Let a be the first term and d be the common difference.

Given: a₁₀ = 41

a₁₆ = 5 a₃

Now, Consider a₁₀ = 41

⇒ a + (10 - 1)d = 41

⇒ a + 9d = 41 ………………….(1)

Consider a₁₆ = 5 a₃

⇒ a + 15d = 5(a + 2d)

⇒ a + 15d = 5a + 10d

⇒ 4a - 5d = 0 ………………….(2)

Now, subtracting equation (2) from 4 times of equation (1), we get,

41d = 164

⇒ d = 4

∴ from equation (2), we get,

4a= 5d

⇒ 4a = 20

⇒ a = 5

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 15 terms is given by:

S₁₅ = [2(5) + (15 - 1)(4)]

= (15/2) × [10 + 56]

= 15 × 33

= 495

∴ S₁₅ = 495

Here, First term = a = 5

Common difference = d = 12 - 5 = 7

No. of terms = 50

∴ last term will be 50^th term.

Using the formula for finding n^th term of an A.P.,

l = a₅₀ = a + (50 - 1) × d

∴ l = 5 + (50 - 1) × 7

⇒ l = 5 + 343 = 348

Now, sum of last 15 terms = sum of first 50 terms - sum of first 35 terms

i.e. sum of last 15 terms = S₅₀ - S₃₅

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 50 terms is given by:

S₅₀ = [2(5) + (50 - 1)(7)]

= 25 × [10 + 343]

= 25 × 353

= 8825

Now, Sum of first 35 terms is given by:

S₃₅ = [2(5) + (35 - 1)(7)]

= (35/2) × [10 + 238]

= (35/2) × 248

= 35 × 124

= 4340

Now, S₅₀ - S₃₅ = 8825 - 4340

= 4485

∴ last term = 348, sum of last 15 terms = 4485

Here, First term = a = 8

Common difference = d = 10 - 8 = 2

No. of terms = 60

∴ last term will be 60^th term.

Using the formula for finding n^th term of an A.P.,

l = a₆₀ = a + (60 - 1) × d

∴ l = 8 + (60 - 1) × 2

⇒ l = 8 + 118 = 126

Now, sum of last 10 terms = sum of first 60 terms - sum of first 50 terms

i.e. sum of last 10 terms = S₆₀ - S₅₀

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 50 terms is given by:

S₅₀ = [2(8) + (50 - 1)(2)]

= 25 × [16 + 98]

= 25 × 114

= 2850

Now, Sum of first 60 terms is given by:

S₆₀ = [2(8) + (60 - 1)(2)]

= 30 × [16 + 118]

= 30 × 248

= 4020

Now, S₆₀ - S₅₀ = 4020 - 2850

= 1170

∴ last term = 126, sum of last 10 terms = 1170

Let a be the first term and d be the common difference.

Given: a₄ + a₈ = 24

and a₆ + a₁₀ = 44

To find: S₁₀

Now, Consider a₄ + a₈ = 24

⇒ a + 3d + a + 7d= 24

⇒ 2a + 10d = 24 ………….(1)

Consider a₆ + a₁₀ = 44

⇒ a + 5d + a + 9d = 44

⇒ 2a + 14d = 44 ………..(2)

Subtracting equation (1) from equation (2), we get,

4d = 20

⇒ d = 5

∴ Common difference = d = 5

Thus from equation (1), we get,

a = - 13

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 10 terms is given by:

S₁₀ = [2(-13) + (10 - 1)(5)]

= 5 × [ - 26 + 45]

= 5 × 19

= 95

∴ S₁₀ = 95

Let a be the first term and d be the common difference.

Given: Sum of first m terms of an AP is given by:

S_m = [2a + (m - 1)d] = 4m² - m

Now, n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (4n² - n) - [4(n - 1)² - (n - 1)]

= (4n² - n) - [4(n² + 1 - 2n) - n + 1]

= 4n² - n - 4n² - 4 + 8n + n - 1

= 8n - 5 …………………(1)

Now, given that a_n = 107

⇒ 8n - 5 = 107

⇒ 8n = 112

⇒ n = 14

For 21^st term of AP, put n = 21 in the value of the nth term in equation (1), we get

a₂₁ = 8 × (21) - 5

⇒ a₂₁ = 168 - 5

= 163

∴ a₂₁ = 163

Let a be the first term and d be the common difference.

Given: Sum of first q terms of an AP is given by:

S_q = [2a + (q - 1)d] = 63q - 3q²

Now, p^th term is given by: a_p = S_p - S_{p - 1}

∴ a_p = (63p - 3p²) - [63(p - 1) - 3(p - 1)²]

= (63p - 3p²) - [63p - 63 - 3p² - 3 + 6p]

= 63p - 3p² - 63p + 63 + 3p² + 3 - 6p

= 66 - 6p …………………(1)

Now, given that a_p = - 60

⇒ 66 - 6p = - 60

⇒ 6p = 126

⇒ p = 21

For 11^th term of AP, put p = 11 in the value of the p^th term in equation (1), we get

a₁₁ = 66 - 6 × (11 )

⇒ a₁₁ = 66 - 66

= 0

∴ a₁₁ = 0

Here, first term = a = - 12

Common difference = d = - 9 - (-12) = 3

Last term is 21.

Now, number of terms in this AP are given as:

21 = a + (n - 1)d

⇒ 21 = - 12 + (n - 1)3

⇒ 21 + 12 = 3n - 3

⇒ 33 + 3 = 3n

⇒ 36 = 3n

⇒ n = 12

If 1 is added to each term, then the new AP will be - 11, - 8, - 5,…, 22.

Here, first term = a = - 11

Common difference = d = - 8 - (-11) = 3

Last term = l = 22.

Number of terms will be the same,

i.e, number of terms = n = 12

∴ Sum of 12 terms of the AP is given by:

S₁₂ = (12/2) × [a + l]

= 6 × [ - 11 + 22]

= 6 × 11

= 66

∴ Sum of 12 terms of the new AP will be 66.

Here, first term = a = 10

Let the Common difference = d

Sum of first 14 terms = S₁₄ = 1505

Now, Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

∴ S₁₄ = [2(10) + (14 - 1)d] = 1505

⇒ 7 × [20 + 13d] = 1505

⇒ [20 + 13d] = 215

⇒ 13d = 195

⇒ d = 15

Now, n^th term is given by:

∴ a_n = a + (n - 1)d

⇒ a₂₅ = 10 + (25 - 1)15

= 10 + (24 × 15)

= 10 + 360

= 370

Here, second term = a₂ = 14

Third term = a₃ = 18

∴ Common difference = a₃ - a₂ = 18 - 14 = 4

Thus first term = a = a₂ - d = 14 - 4 = 10

Now, Sum of first n terms of an AP is given by

S_n = [2a + (n - 1)d]

∴ Sum of first 51 terms is given by:

S₅₁ = [2(10) + (51 - 1)(4)]

= (51/2) × [20 + 200]

= (51/2) × 220

= (51) × 110

= 5610

∴ S₅₁ = 5610

Number of trees planted by one section of class 1^st = 2

Now, there are 2 sections, ∴ Number of trees planted by class 1^st = 4

Number of trees planted by one section of class 2^nd = 4

Now, there are 2 sections, ∴ Number of trees planted by class 2^nd = 8

This will follow up to class 12^th and we will obtain an AP as

4, 8, 12, … upto 12 terms.

Now, Total number of trees planted by the students = 4 + 8 + 12 + … upto 12 terms.

∴ In this Arithmetic series, first term = a = 4

Common difference = d = 4

Now, S₁₂ = (12/2)[2a + (12 - 1)d]

= 6[2(4) + 11(4)]

= 6 × [8 + 44]

= 6 × 52

= 312

∴ Total number of trees planted by the students = 312

Values shown in the question are care and awareness about conservation of nature and environment.

To pick the first potato, the competitor has to run 5 m to reach the potato and 5 m to run back to the bucket.

∴ Total distance covered by the competitor to pick first potato = 2 × (5) = 10 m

To pick the second potato, the competitor has to run (5 + 3) m to reach the potato and (5 + 3) m to run back to the bucket.

∴ Total distance covered by the competitor to pick second potato = 2 × (5 + 3) = 16 m

To pick the third potato, the competitor has to run (5 + 3 + 3) m to reach the potato and (5 + 3 + 3) m to run back to the bucket.

∴ Total distance covered by the competitor to pick third potato = 2 × (5 + 3 + 3) = 22 m

This will continue and we will get a sequence of distance as 10, 16, 22,… upto 10 terms (as there are 10 potatoes to pick).

Total distance covered by the competitor to pick all the 10 potatoes = 10 + 16 + 22 + … upto 10 terms.

This forms an Arithmetic series with first term = a = 10

and Common difference = d = 6

Number of terms = n = 10

Now, S₁₀ = (10/2)[2a + (10 - 1)d]

= 5 × [2(10) + 9(6)]

= 5 × [20 + 54]

= 5 × 74

= 370

∴ Total distance covered by the competitor = 370 m

To water the first tree, the gardener has to cover 10 m to reach the tree and 10 m to go back to the tank.

∴ Total distance covered by the gardener to water first tree = 2 × (10) = 20 m

To water the second tree, the gardener has to cover (10 + 5) m to reach the tree and (10 + 5) m to go back to the tank.

∴ Total distance covered by the gardener to water second tree = 2 × (10 + 5) = 30 m

To water the third tree, the gardener has to cover (10 + 5 + 5) m to reach the tree and (10 + 5 + 5) m to go back to the tank.

∴ Total distance covered by the gardener to water third tree = 2 × (10 + 5 + 5) = 40 m

This will continue and we will get a sequence of distance as 20, 30, 40,… upto 25 terms (as there are 25 trees to be watered).

Total distance covered by the gardener to water all 25 trees = 20 + 30 + 40 + … upto 25 terms.

This forms an Arithmetic series with first term = a = 20

and Common difference = d = 10

Number of terms = n = 25

Now, S₂₅ = (25/2)[2a + (25 - 1)d]

= (25/2) × [2(20) + 24(10)]

= (25/2) × [40 + 240]

= (25/2) × 280

= 25 × 140

= 3500

∴ Total distance covered by the gardener = 3500 m

Let the first prize be Rs. x. Thus each succeeding prize is Rs. 20 less than the preceding prize.

∴ Second prize, third prize, …, seventh prize be Rs. (x - 20), (x - 40) , …, (x - 120).

This forms an AP as x, x - 20, …, x - 120.

Here, first term = x

Common difference = x - 20 - x = - 20

Total number of terms = 7

Since, Total sum of prize amount = 700.

∴ Sum of all the terms = 700

Now, sum of first n terms of an AP is given by:

S_n = [2a + (n - 1)d]

∴ Sum of 7 terms of an AP is given by:

S₇ = [2a + (7 - 1)d] = 700

⇒ [2x + (7 - 1)(-20)] = 700

⇒ 7[2x - 120] = 1400

⇒ 2x - 120 = 200

⇒ x - 60 = 100

⇒ x = 160

Thus, the prizes are as Rs. 160, Rs.140, Rs.120, Rs. 100, Rs. 80, Rs. 60, Rs. 40.

Let the amount of money the man saved in first month = Rs. x

Now, the amount of money he saved in second month = Rs.(x + 100)

The amount of money he saved in third month = Rs.(x + + 100 + 100)

This will continue for 10 months.

∴ We get a an AP as x , x + 100, x + 200,… up to 10 terms.

Here, first term = x

Common difference = d = 100

Number of terms = n = 10

Total amount of money saved by the man = x + (x + 100) + (x + 200) + … up to 10 terms. = Rs. 33000 (given)

∴ Sum of 10 terms of the Arithmetic Series = 33000

⇒ S₁₀ =33000

⇒ (10/2) × [2a + (10 - 1)d] =33000

⇒ (10/2) × [2(x) + 9(100)] =33000

⇒ 5 × [2x + 900] =33000

⇒ 2x + 900 =6600

⇒ 2x = 6600 - 900

⇒ 2x = 5700

⇒ x = 2850

∴ Amount of money saved by the man in first month = Rs. 2850

Let the first installment = Rs. x

Since the instalments form an arithmetic series, therefore let the common difference = d

Now, amount paid in 30 installments = two - third of the amount = (2/3) × (36000) = Rs. 24000

∴ Total amount paid by the man in 30 installments = 24000

Let S_n be that amount paid in 30 installments.

∴ S₃₀ = 24000

⇒ (30/2) × [2x + (30 - 1)d] = 24000

⇒ 15 × [2x + 29d] = 24000

⇒ 2x + 29d = 1600 …………..(1)

Now, Total sum of the amount = 36000

∴ S₄₀ = 36000

⇒ (40/2) × [2x + (40 - 1)d] = 36000

⇒ 20 × [2x + 39d] = 36000

⇒ 2x + 39d = 1800 ………..(2)

Subtracting equation (1) from equation (2), we get:

10d = 200

⇒ d = 20

∴ from equation (1), we get

x = 1/2(1600 - 29d)

= 1/2 (1600 - 580)

= 1/2 (1020)

= 510

Therefore the amount of first installment = Rs. 510

Penalty for delay for first day = Rs. 200

Penalty for delay for second day = Rs. 250

Penalty for delay for third day = Rs. 300

Penalty for each succeeding day is Rs. 50 more than for the preceding day.

∴ The amount of penalties are in AP with common difference

= d = Rs.50

Also, number of days in delay of the work = 30 days

Thus the penalties are 200, 250, 300, … up to 30 terms

Thus the amount of money paid by the contractor is 200 + 250 + 300 + … up to 30 terms

Here, first term = a = 200

Common difference = d = 50

Number of terms = n = 30

∴ The sum = S₃₀ = (30/2) × [2(200) + (30 - 1)(50)]

= 15 × [400 + 1450]

= 15 × 1850

= 27750

Thus the total amount of money paid by the contractor = Rs. 27750

Common difference = T₂ - T₁ =

= = - 1

Common difference = T₂ - T₁ =

Here, first term = √7

Common difference = √28 - √7 = 2√7 - √7 = √7

Next term = T₄ = T₃ + d

= √63 + √7

= 3√7 + √7

= 4√7

= √112

Here, first term = a = 4

Last term = l = 28

Number of terms = n = 5

∴ l = a + (n - 1)d

⇒ 28 = 4 + (5 - 1)d

⇒ 28 - 4 = 4d

⇒ 4d = 24

⇒ d = 6

Therefore x₃ = a + 3d

= 4 + 3(6)

= 4 + 18

= 22

Given: n^th term = 2n + 1

∴ a_n = 2n + 1

Put n= 1, a₁ = 3

Put n= 2, a₂ = 5

Put n= 3, a₃ = 7

Now, sum of first three terms = 3 + 5 + 7 = 15

Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 3n² + 6n

Then n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (3n² + 6n) - [3(n - 1)² + 6(n - 1)]

= (3n² + 6n) - [3(n² + 1 - 2n) + 6n - 6]

= 3n² + 6n - 3n² - 3 + 6n - 6n + 6

= 3 + 6n

Now, common difference = d = a_n - a_{n - 1}

= 3 + 6n - [3 + 6(n - 1)]

= 3 + 6n - 3 - 6n + 6

= 6

∴ Common difference = 6

Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 5n - n²

Then n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (5n - n²) - [5(n - 1) - (n - 1)²]

= (5n - n²) - [5n - 5 - (n² + 1 - 2n)]

= 5n - n² - 5n + 5 + n² + 1 - 2n

= 6 - 2n

Let S_n denotes the sum of first n terms of an AP.

Sum of first n terms = S_n = 4n² + 2n

Then n^th term is given by: a_n = S_n - S_{n - 1}

∴ a_n = (4n² + 2n) - [4(n - 1)² + 2(n - 1)]

= (4n² + 2n) - [4(n² + 1 - 2n) + 2n - 2]

= 4n² + 2n - 4n² - 4 + 8n - 2n + 2

= 8n - 2

Let a be the first term and d be the common difference.

Given: a₇ = - 1

a₁₆ = 17

Now, Consider a₇ = - 1

⇒ a + 6d = - 1 …………(1)

Consider a₁₆ = 17

⇒ a + 15d = 17 …………(2)

Subtract equation (1) from (2), we get,

9d = 18

⇒ d = 2

∴ Common difference = d = 2

Now, from equation (1), we get,

a = - 1 - 6d

= - 1 - 6(2)

= - 13

Now, n^th term of the AP is given by

a_n = a + (n - 1)d

= - 13 + (n - 1)2

= 13 + 2n - 2

= 2n - 15

Let a be the first term and d be the common difference.

Given: a₅ = - 3

Common difference = d = - 4

Now, Consider a₅ = - 3

⇒ a + 4d = - 3

⇒ a + 4(-4) = - 3

⇒ a - 16 = - 3

⇒ a = 16 - 3

⇒ a = 13

Now, Sum of first n terms of an AP is

S_n = [2a + (n - 1)d]

∴ Sum of first 10 terms is given by:

S₁₀ = [2(13) + (10 - 1)(-4)]

= 5[26 - 36]

= 5 × (-10)

= - 50

Let a be the first term and d be the common difference.

Given: a₅ = 20

a₇ + a₁₁ = 64

Now, Consider a₅ = 20

⇒ a + 4d = 20 ……………(1)

Consider a₇ + a₁₁ = 64

⇒ a + 6d + a + 10d = 64

⇒ 2a + 16d = 64 …………(2)

Subtract twice of equation (1) from (2), we get,

8d = 24

⇒ d = 3

Let a be the first term and d be the common difference.

Given: a₁₃ = 4(a₃)

a₅ = 16

To find: Sum of first ten terms.

Now, Consider a₁₃ = 4a₃

⇒ a + 12d = 4[a + 2d]

⇒ a + 12d = 4a + 8d

⇒ 3a = 4d ………. (1)

Consider a₅ = a + (5 - 1)d = 16

⇒ a + 4d = 16

⇒ a + 3a = 16 (from equation (1))

⇒ 4a = 16

⇒ a = 4 ………. (2)

∴ d = 3

Sum of n terms of an arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 10 terms of the arithmetic series is given by:

∴ S₁₀ = [2(4) + (10 - 1)(3)]

= 5 × [8 + 27]

= 5 × 35

= 175

Here, first term = 5

Common difference = 12 - 5 = 7

Given that there are 50 terms in the AP.

To find: Last term, i.e. 50^th term = a₅₀

Since a_n = a + (n - 1)d

∴ a₅₀ = 5 + (50 - 1)7

= 5 + (49) × 7

= 5 + 343

= 348

Sum of first 20 odd natural numbers is 1 + 3 + 5 + 7 + … + 39.

This forms an arithmetic series with first term = a = 1

and common difference = d = 2

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 20 terms of this arithmetic series is given by:

∴ S₂₀ = [2(1) + (20 - 1)(2)]

= 10 [2 + 38]

= 10 × 40

= 400

First 40 positive integers divisible by 6 are 6, 12, 18, …, 240.

Sum of these numbers forms an arithmetic series 6 + 12 + 18 + … + 240.

Here, first term = a = 6

Common difference = d = 6

Sum of n terms of this arithmetic series is given by:

S_n = [2a + (n - 1)d]

Therefore sum of 40 terms of this arithmetic series is given by:

∴ S₄₀ = [2(6) + (40 - 1)(6)]

= 20 [12 + 234]

=20 × 246

= 4920

The two digit numbers divisible by 3 are 12, 15, 18, 21, …., 99.

This forms an AP with first term a = 12

and common difference = d = 3

Last term is 99.

Now, number of terms in this AP are given as:

99 = a + (n - 1)d

⇒ 99 = 12 + (n - 1)3

⇒ 99 - 12 = 3n - 3

⇒ 87 + 3 = 3n

⇒ 90 = 3n

⇒ n = 30

There are 30 two - digit numbers that are divisible by 3.

The three digit numbers divisible by 9 are 108, 117, 126, …., 999.

This forms an AP with first term a = 108

and common difference = d = 9

Last term is 999.

Now, number of terms in this AP are given as:

999 = a + (n - 1)d

⇒ 999 = 108 + (n - 1)9

⇒ 999 - 108 = 9n - 9

⇒ 891 + 9 = 9n

⇒ 900 = 9n

⇒ n = 100

There are 100 three - digit numbers that are divisible by 9.