Arithmetic Progressions (ap)

Given: t_n = 3n + 1

Taking n = 1, we get

t₁ = 3(1) + 1 = 3 + 1 = 4

Taking n = 2, we get

t₂ = 3(2) + 1 = 6 + 1 = 7

Taking n = 3, we get

t₃ = 3(3) + 1 = 9 + 1 = 10

Hence, the first three terms are 4, 7 and 10.

Given: t_n = 2ⁿ

Taking n = 1, we get

t₁ = 2¹ = 2

Taking n = 2, we get

t₂ = 2² = 2 × 2 = 4

Taking n = 3, we get

t₃ = 2³ = 2 × 2 × 2 = 8

Hence, the first three terms are 2, 4 and 8.

Given: t_n = n² + 1

Taking n = 1, we get

t₁ = (1)² + 1 = 1 + 1 = 2

Taking n = 2, we get

t₂ = (2)² + 1 = 4 + 1 = 5

Taking n = 3, we get

t₃ = (3)² + 1 = 9 + 1 = 10

Hence, the first three terms are 2, 5 and 10.

Given: t_n = n(n+2)

Taking n = 1, we get

t₁ = (1)(1+2) = (1)(3) = 3

Taking n = 2, we get

t₂ = (2)(2+2) = (2)(4) = 8

Taking n = 3, we get

t₃ = (3)(3+2) = (3)(5) = 15

Hence, the first three terms are 3, 8 and 15.

Given: t_n = 2n + 5

Taking n = 1, we get

t₁ = 2(1) + 5 = 2 + 5 = 7

Taking n = 2, we get

t₂ = 2(2) + 5 = 4 + 5 = 9

Taking n = 3, we get

t₃ = 2(3) + 5 = 6 + 5 = 11

Hence, the first three terms are 7, 9 and 11.

Given:

Taking n = 1, we get

Taking n = 2, we get

Taking n = 3, we get

Hence, the first three terms are

Given:

Now, we have to find t₁ and t₂.

So, in t₁, n = 1

Now, t₂, n = 2

Given:

So,

Given: t_n = (n – 1)(2 – n)(3+n)

So, t₂₀ = (20 – 1)(2 – 20)(3+20)

= (19)(-18)(23)

= -7866

Given:

So, [given: t₁ = 3]

and

Given: t₂ = 2 and t_n = t_n-1 + 1

Now, we have to find next three terms i.e. t₃, t₄ and t₅

Taking n = 3, we get

t₃ = t_3-1 + 1

= t₂ + 1

= 2 + 1 [given: t₂ = 2]

t₃ = 3 …(i)

Taking n = 4, we get

t₄ = t_4-1 + 1

= t₃ + 1

= 3 + 1 [from (i)]

t₄ = 4 …(ii)

Taking n = 5, we get

t₅ = t_5-1 + 1

= t₄ + 1

= 4 +1

t₅ = 5 [from (ii)]

Hence, the next three terms are 3, 4 and 5.

Given: t₁ = 3 and t_n = 3t_n-1 + 2

Now, we have to find next three terms i.e. t₂, t₃ and t₄

Taking n = 2, we get

t₂ = 3t_2-1 + 2

= 3t₁ + 2

= 3(3) + 2 [given: t₁ = 3]

t₃ = 9 + 2

t₃ = 11 …(i)

Taking n = 3, we get

t₃ = 3t_3-1 + 2

= 3t₂ + 2

= 3(11) + 2 [from (i)]

= 33 + 2

t₃ = 35 …(ii)

Taking n = 4, we get

t₄ = 3t_4-1 + 2

= 3t₃ + 2

= 3(35) +2

t₅ = 105 + 2

t₅ = 107 [from (ii)]

Hence, the next three terms are 11, 35 and 107.

Given: a = 1 and d = 1

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is 1 and the common difference d is 1, then the first four terms of the AP is

1, (1+1), (1+2×1), (1+3×1)

⇒ 1, 2, 3, 4

Hence, the first four terms of the A.P. is 1, 2, 3 and 4.

Given: a = 3 and d = 0

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is 3 and the common difference d is 0, then the first four terms of the AP is

3, (3+0), (3+2×0), (3+3×0)

⇒ 3, 3, 3, 3

Hence, first four terms of the A.P. is 3, 3, 3 and 3.

Given: a = 10 and d = 10

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is 10 and the common difference d is 10, then the first four terms of the AP is

10, (10+10), (10+2×10), (10+3×10)

⇒ 10, (20), (10+20), (10+30)

⇒ 10, 20, 30, 40

Hence, first four terms of the A.P. is 10, 20, 30 and 40.

Given: a = -2 and d = 0

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is -2 and the common difference d is 0, then the first four terms of the AP is

-2, (-2+0), (-2+2×0), (-2+3×0)

⇒ -2, -2, -2, -2

Hence, the first four terms of the A.P. is -2, -2, -2 and -2.

Given: a = 100 and d = -30

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is 100 and the common difference d is -30, then the first four terms of the AP is

100, (100+(-30)), (100+2×(-30)),(100+3×(-30))

⇒ 100, (100 – 30), (100 – 60), (100 – 90)

⇒ 100, 70, 40, 10

Hence, first four terms of the A.P. is 100, 70, 40 and 10.

Given: a = -1 and d

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is 1 and the common difference d is , then the first four terms of the AP is

⇒

Hence, first four terms of the A.P. is .

Given: a = -7 and d = -7

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is -7, and the common difference d is -7, then the first four terms of the AP is

-7, (-7+(-7)), (-7+2×(-7)), (-7+3×(-7))

⇒ -7, (-7 – 7), (-7 – 14), (-7 – 21)

⇒ -7, -14, -21, -28

Hence, the first four terms of the A.P. is -7, -14, -21 and -28.

Given: a = 1 and d = 0.1

The general form of an A.P is a, a+d, a+2d, a+3d,…

So, the first term a is 1, and the common difference d is 0.1, then the first four terms of the AP is

1, (1+0.1), (1+2×(0.1)), (1+3×(0.1))

⇒ 1, 1.1, 1.2, 1.3

Hence, the first four terms of the A.P. is 1, 1.1, 1.2 and 1.3.

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: 6, 3, 0, -3, . . .

a₂ – a₁ = 3 – 6 = -3

a₃ – a₂ = 0 – 3 = -3

a₄ – a₃ =-3 – 0 = -3

Here, the difference of any two consecutive terms in each case is -3.

So, the given list is an AP whose first term a is 6, and common difference d is -3.

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: - 3.1, - 3.0, - 2.9, - 2.8, ...

a₂ – a₁ = -3.0 – (-3.1) = -3.0 + 3.1 = 0.1

a₃ – a₂ = -2.9 – (-3.0) = -2.9 + 3.0 = 0.1

a₄ – a₃ = -2.8 – (-2.9) = -2.8 + 2.9 = 0.1

Here, the difference of any two consecutive terms in each case is 0.1. So, the given list is an AP whose first term a is -3.1 and common difference d is 0.1.

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: 147, 148, 149, 150, ...

a₂ – a₁ = 148 – 147 = 1

a₃ – a₂ = 149 – 148 = 1

a₄ – a₃ = 150 – 149 = 1

Here, the difference of any two consecutive terms in each case is -1. So, the given list is an AP whose first term a is 147 and common difference d is 1.

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: - 5, - 1, 3, 7, ...

a₂ – a₁ = -1 – (-5) = -1 + 5 = 4

a₃ – a₂ = 3 – (-1) = 3 + 1 = 4

a₄ – a₃ = 7 – 3 = 4

Here, the difference of any two consecutive terms in each case is -4. So, the given list is an AP whose first term a is -5 and common difference d is 4.

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers: 3, 1, - 1, - 3, ...

a₂ – a₁ = 1 – 3 = -2

a₃ – a₂ =-1 – 1 = -1 - 1 = -2

a₄ – a₃ =-3 – (-1) = -3 + 1= -2

Here, the difference of any two consecutive terms in each case is --2. So, the given list is an AP whose first term a is 3 and common difference d is -2.

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers:

Here, the difference of any two consecutive terms in each case is . So, the given list is an AP whose first term a is 2 and common difference d is .

In general, for an AP a₁, a₂, . . . .,a_n, we have

d = a_k+1 - a_k

where a_k+1 and a_k are the (k+1)^th and k^th terms respectively.

For the list of numbers:

Here, the difference of any two consecutive terms in each case is -1. So, the given list is an AP whose first term a is and common difference d is -1.

We have,

a₂ – a₁ = -1 – 1 = -2

a₃ – a₂ =-3 – (-1) = -3 + 1 = -2

a₄ – a₃ =-5 – (-3) = -5 + 3= -2

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = -2

Now, we have to find the next three terms.

We have a₁ = 1, a₂ = -1, a₃ = -3 and a₄ = -5

Now, we will find a₅, a₆ and a₇

So, a₅ = -5 + (-2) = -5 – 2 = -7

a₆ = -7 + (-2) = -7 – 2 = -9

and a₇ = -9 + (-2) = -9 – 2 = -11

Hence, the next three terms are -7, -9 and -11

We have,

a₂ – a₁ = 4 – 2 = 2

a₃ – a₂ = 8 – 4 = 4

a₄ – a₃ = 16 – 8 = 8

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers do not form an AP.

We have,

a₂ – a₁ = 2 – (-2) = 2 + 2 = 4

a₃ – a₂ = -2 – 2 = -4

a₄ – a₃ = 2 – (-2) = 2 + 2 = 4

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers do not form an AP.

We have,

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = 0

Now, we have to find the next three terms.

We have

Now, we will find a₅, a₆ and a₇

So,

and

Hence, the next three terms are

We have,

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference

Now, we have to find the next three terms.

We have

Now, we will find a₅, a₆ and a₇

So,

and

Hence, the next three terms are

We have,

a₂ – a₁ = -4 – 0 = -4

a₃ – a₂ =-8 – (-4) = -8 + 4 = -4

a₄ – a₃ =-12 – (-8) = -12 + 8= -4

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = -4

Now, we have to find the next three terms.

We have a₁ = 0, a₂ = -4, a₃ = -8 and a₄ = -12

Now, we will find a₅, a₆ and a₇

So, a₅ = -12 + (-4) = -12 – 4 = -16

a₆ = -16 + (-4) = -16 – 4 = -20

and a₇ = -20 + (-4) = -20 – 4 = -24

Hence, the next three terms are -16, -20 and -24

We have,

a₂ – a₁ = 10 – 4 = 6

a₃ – a₂ = 16 – 10 = 6

a₄ – a₃ = 22 – 16 = 6

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = 6

Now, we have to find the next three terms.

We have a₁ = 4, a₂ = 10, a₃ = 16 and a₄ = 22

Now, we will find a₅, a₆ and a₇

So, a₅ = 22 + 6 = 28

a₆ = 28 + 6 = 34

and a₇ = 34 + 6 = 40

Hence, the next three terms are 28, 34 and 40

We have,

a₂ – a₁ = 2a – a = a

a₃ – a₂ = 3a – 2a = a

a₄ – a₃ = 4a – 3a = a

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = a

Now, we have to find the next three terms.

We have a₁ = a, a₂ = 2a, a₃ = 3a and a₄ = 4a

Now, we will find a₅, a₆ and a₇

So, a₅ = 4a + a = 5a

a₆ = 5a + a = 6a

and a₇ = 6a + a = 7a

Hence, the next three terms are 5a, 6a and 7a

We have,

a₂ – a₁ = -3.2 – (-1.2) =-3.2 + 1.2 = -2.0

a₃ – a₂ = -5.2 – (-3.2) = -5.2 + 3.2 = -2.0

a₄ – a₃ = -7.2 – (-5.2) = -7.2 + 5.2 = -2.0

i.e. a_k+1 – a_k is the same every time.

So, the given list of numbers forms an AP with the common difference d = -2

Now, we have to find the next three terms.

We have a₁ = -1.2, a₂ = -3.2, a₃ = -5.2 and a₄ = -7.2

Now, we will find a₅, a₆ and a₇

So, a₅ = -7.2 + (-2) = -7.2 – 2.0 = -9.2

a₆ = -9.2 + (-2) = -9.2 – 2.0 = -11.2

and a₇ = -11.2 + (-2) = -11.2 – 2.0 = -13.2

Hence, the next three terms are -9.2, -11.2 and -13.2

We have,

a₂ – a₁ = √12 - √3 = 2√3 - √3 = √3

a₃ – a₂ = √48 - √12 = 4√3 - 2√3 = 2√3

a₄ – a₃ = √192 - √48 = 8√3 - 4√3 = 4√3

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers do not form an AP.

We have,

a₂ – a₁ = a² – a = a (a – 1)

a₃ – a₂ = a³ – a² = a²(a – 1)

a₄ – a₃ = a⁴ – a³ = a³(a – 1)

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers does not form an AP.

We have,

a₂ – a₁ = 3 – 1 = 2

a₃ – a₂ = 9 – 3 = 6

a₄ – a₃ = 27 – 9 = 18

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers does not form an AP.

We have,

a₂ – a₁ = 2² – (1)² = 4 – 1 = 3

a₃ – a₂ = 3² – (2)² = 9 – 4 = 5

a₄ – a₃ = 4² – (3)² = 16 – 9 =7

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers does not form an AP.

We have,

a₂ – a₁ = 5² – (1)² = 25 – 1 = 24

a₃ – a₂ = 7² – (5)² = 49 – 25 = 24

a₄ – a₃ = 7² – (7)² = 0

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers do not form an AP.

We have,

a₂ – a₁ = 3² – (1)² = 9 – 1 = 8

a₃ – a₂ = 5² – (3)² = 25 – 9 = 16

a₄ – a₃ = 7² – (5)² = 49 – 25 =24

i.e. a_k+1 – a_k is not same every time.

So, the given list of numbers does not form an AP.

Salary for the 1^st year = Rs. 8000

and according to the question,

There is an annual increment of Rs. 500

⇒ The salary for the 2^nd year = Rs. 8000 + 500 = Rs.8500

Now, again there is an increment of Rs. 500

⇒ The salary for the 3^rd year = Rs. 8500 +500 = Rs. 9000

Therefore, the series is

8000 , 8500 , 9000 , …

Difference between 2^nd term and 1^st term = 8500 – 8000 = 500

Difference between 3^rd term and 2^nd term = 9000 – 8500 = 500

Since, the difference is same.

Hence, salary in successive years are in AP with common difference d = 500 and first term a is 8000.

Taxi fare for 1km = 15

According to question, Rs. 8 for each additional km

⇒ Taxi fare for 2km = 15 + 8 =23

and Taxi fare for 3km = 23 + 8 =31

Therefore, series is

15, 23, 31 ,…

Difference between 2^nd and 1^st term = 23 – 15 = 8

Difference between 3^rd and 2^nd term = 31 – 23 = 8

Since, difference is same.

Hence, the taxi fare after each km form an AP with the first term, a = Rs. 15 and common difference, d = Rs. 8

The length of the bottom rung = 45cm

According to the question,

Length of rungs decreases by 2cm from bottom to top. The lengths (in cm) of the 1^st, 2^nd, 3^rd, … from the bottom to top respectively are 45, 43, 41, …

Difference between 2^nd and 1^st term = 43 – 45 = -2

Difference between 3^rd and 2^nd term = 41 – 43 = -2

Since, the difference is same.

Hence, the length of the rungs form an AP with a = 45cm and d = -2 cm.

Original Amount = Rs. 10,000

Interest earned in first year = 10,000 × 8%

= Rs 800

Total amount outstanding after one year = Rs 10000 + 800

= Rs 10800

Now, interest earned in 2^nd year

Total amount outstanding after 2^nd year = Rs10800 + 864

= Rs 11664

Interest earned in 3^rd year

= Rs 933.12

Total amount outstanding after 3^rd year = Rs 11664 + 933.12

= Rs 12597.12

Therefore, the series is

10800, 11664, 12597.12,…

Difference between second and first term = 11664 – 10800

= 864

Difference between third and second term = 12597.12 – 11664

= 933.12

Since the difference is not same

Therefore, it doesn’t form an AP.

The money saved by Sudha in the first year = Rs. 100

According to the question,

Sudha increased the amount by Rs. 50 every year

⇒ The money saved by Sudha in a 2^nd year = Rs. 100 +50

= Rs. 150

The money saved by Sudha in a 3^rd year = Rs. 150 + 50

= Rs. 200

Therefore, the series is

100, 150, 200, 250,…

Difference in the 2^nd term and 1^st term = 150 – 100 = 50

Difference in the 3^rd term and 2^nd term = 200 – 150 = 50

Since the difference is the same.

Therefore, the money saved by Sudha in successive years form an AP with a = Rs 100 and d =Rs 50

Assuming that no rabbit dies,

the number of pairs of rabbits at the start of the 1^st month = 1

the number of pairs of rabbits at the start of the 2^nd month = 1

the number of pairs of rabbits at the start of the 3^rd month = 2

the number of pairs of rabbits at the start of the 4^th month = 3

the number of pairs of rabbits at the start of the 5^th month = 5

Therefore, the series is

1, 1, 2, 3, 5, 8,…

Difference between 2^nd and 1^st term = 1 – 1 = 0

Difference between 3^rd and 2^nd term = 2 – 1 = 1

Since, the difference is not same.

Therefore, the number of pair of rabbits in successive months are 1,1,2,3,5,8,… and they don’t form an AP.

Let the initial investment be I,

After one year it increases by I,

So the investment becomes, I + I

= I

At the end of 2^nd year it again increase to I,

So the investment becomes, (I + I + I )

=( I + I)

= I

At the end of the 3rd year it again increases to I

So the investment becomes (I + I + I + I )

= I + I + I

= I+ I

= I

Therefore the series is:

I, I, I, I,………

Now difference between 2^nd and 1^st term is I – I = I

difference between 3^rd and 2^nd term is I I = I

Since the difference is same,

Hence the obtained series is an A.P.

Given: 1, 6, 11, 16, …

Here, a = 1

d = a₂ – a₁ = 6 – 1 = 5

and n = 16

We have,

t_n = a + (n – 1)d

So, t₁₆ = 1 + (16 – 1)5

= 1 + 15×5

t₁₆ = 1 +75

t₁₆ = 76

Given: a = 3 , d = 2

To find: t_n and t₁₀

We have,

t_n = a + (n – 1)d

t_n = 3 + (n – 1) 2

= 3 + 2n – 2

t_n = 2n + 1

Now, n = 10

So, t₁₀ = 3 + (10 – 1)2

= 3 + 9×2

t₁₀ = 3 +18

t₁₀ = 21

Given:

Here, a = –3

and n = 10

We have,

t_n = a + (n – 1)d

So,

Given: a = 21 , d = –5

To find: t_n and t₂₅

We have,

t_n = a + (n – 1)d

t_n = 21 + (n – 1)(–5)

= 21 – 5n + 5

t_n = 26 – 5n

Now, n = 25

So, t₂₅ = 21 + (25 – 1)(–5)

= 21 + 24 × (–5)

t₂₅ = 21 – 120

t₂₅ = –99

Given: 10, 5, 0, –5, –10,…

To find: 10^th term i.e. t₁₀

Here, a = 10

d = a₂ – a₁ = 5 – 10 = –5

and n = 10

We have,

t_n = a + (n – 1)d

t₁₀ = 10 + (10 – 1)(–5)

= 10 + 9 × –5

t₁₀ = 10 – 45

t₁₀ = –35

Therefore, the 10^th term of the given is –35.

Given:

Here,

and n = 10

We have,

t_n = a + (n – 1)d

Therefore, the 10^th term of the given AP is

Given: 2, 5, 8, 11, …

Here, a = 2

d = a₂ – a₁ = 5 – 2 = 3

and n = 20

We have,

t_n = a + (n – 1)d

t₂₀ = 2 + (20 – 1)(3)

t₂₀ = 2 + 19 × 3

= 2 + 57

t₂₀ = 59

Now, n = 25

t₂₅ = 2 + (25 – 1)(3)

t₂₅ = 2 + 24 × 3

t₂₅ = 2 + 72

t₂₅ = 74

The sum of 20^th and 25^th terms of AP = t₂₀ + t₂₅ = 59 + 74 = 133

Here, a = 6 , d = 3 – 6 = –3 and l = –36

where l = a + (n – 1)d

⇒ –36 = 6 + (n – 1)(–3)

⇒ –36 = 6 –3n + 3

⇒ –36 = 9 – 3n

⇒ –36 – 9 = –3n

⇒ –45 = –3n

Hence, the number of terms in the given AP is 15

Here,

And

We have,

l = a + (n – 1)d

⇒ 15 = n – 1

⇒ n = 16

Hence, the number of terms in the given AP is 16.

Here, a = 3, d = 7 – 3 = 4 and l = 399

To find : n and 20^th term from the end

We have,

l = a + (n – 1)d

⇒ 399 = 3 + (n – 1) × 4

⇒ 399 – 3 = 4n – 4

⇒ 396 + 4 = 4n

⇒ 400 = 4n

⇒ n = 100

So, there are 100 terms in the given AP

Last term = 100^th

Second Last term = 100 – 1 = 99^th

Third last term = 100 – 2 = 98^th

And so, on

20^th term from the end = 100 – 19 = 81^st term

The 20^th term from the end will be the 81^st term.

So, t₈₁ = 3 + (81 – 1)(4)

t₈₁ = 3 + 80 × 4

t₈₁ = 3 + 320

t₈₁ = 323

Hence, the number of terms in the given AP is 100, and the 20^th term from the last is 323.

Here, a = 5, d = 9 – 5 = 4 and a_n = 81

To find : n

We have,

a_n = a + (n – 1)d

⇒ 81 = 5 + (n – 1) × 4

⇒ 81 = 5 + 4n – 4

⇒ 81 = 4n + 1

⇒ 80 = 4n

⇒ n = 20

Therefore, the 20^th term of the given AP is 81.

Here, a = 14, d = 9 – 14 = –5 and a_n = –41

To find : n

We have,

a_n = a + (n – 1)d

⇒ –41 = 14 + (n – 1) × (–5)

⇒ –41 = 14 – 5n + 5

⇒ –41 = 19 – 5n

⇒ – 41 – 19 = –5n

⇒ –60 = –5n

⇒ n = 12

Therefore, the 12^th term of the given AP is –41.

Here, a = 3, d = 8 – 3 = 5 and a_n = 88

To find : n

We have,

a_n = a + (n – 1)d

⇒ 88 = 3 + (n – 1) × (5)

⇒ 88 = 3 + 5n – 5

⇒ 88 = –2 + 5n

⇒88 + 2 = 5n

⇒ 90 = 5n

⇒ n = 18

Therefore, the 18^th term of the given AP is 88.

Here,

and a_n = 3

We have,

a_n = a + (n – 1)d

⇒ 13 = n – 1

⇒ n = 14

Therefore, the 14^th term of a given AP is 3.

Here, a = 3, d = 8 – 3 = 5 and a_n = 248

To find : n

We have,

a_n = a + (n – 1)d

⇒ 248 = 3 + (n – 1) × (5)

⇒ 248 = 3 + 5n – 5

⇒ 248 = –2 + 5n

⇒ 248 + 2 = 5n

⇒ 250 = 5n

⇒ n = 50

Therefore, the 50^th term of the given AP is 248.

Here, a = 17, d = 14 – 17 = –3 and l = –40

where l = a + (n – 1)d

Now, to find the 6^th term from the end, we will find the total number of terms in the AP.

So, –40 = 17 + (n – 1)(–3)

⇒ –40 = 17 –3n + 3

⇒ –40 = 20 – 3n

⇒ –60 = –3n

⇒ n = 20

So, there are 20 terms in the given AP.

Last term = 20^th

Second last term = 20 – 1 = 19^th

Third last term = 20 – 2 = 18^th

And so, on

So, the 6^th term from the end = 20 – 5 = 15^th term

So, a_n = a + (n – 1)d

⇒ a₁₅ = 17 + (15 – 1)(–3)

⇒ a₁₅ = 17 + 14 × –3

⇒ a₁₅ = 17 – 42

⇒ a₁₅ = –25

Here, a = 7, d = 10 – 7 = 3 and l = 184

where l = a + (n – 1)d

Now, to find the 8^th term from the end, we will find the total number of terms in the AP.

So, 184 = 7 + (n – 1)(3)

⇒ 184 = 7 + 3n – 3

⇒ 184 = 4 + 3n

⇒ 180 = 3n

⇒ n = 60

So, there are 60 terms in the given AP.

Last term = 60^th

Second last term = 60 – 1 = 59^th

Third last term = 60 – 2 = 58^th

And so, on

So, the 8^th term from the end = 60 – 7 = 53^th term

So, a_n = a + (n – 1)d

⇒ a₅₃ = 7 + (53 – 1)(3)

⇒ a₅₃ = 7 + 52 × 3

⇒ a₅₃ = 7 + 156

⇒ a₅₃ = 163

Here, a = 6 , d = 10 – 6 = 4 and l = 174

where l = a + (n – 1)d

⇒ 174 = 6 + (n – 1)(4)

⇒ 174 = 6 + 4n – 4

⇒ 174 = 2 + 4n

⇒ 174 – 2 = 4n

⇒ 172 = 4n

Hence, the number of terms in the given AP is 43

Here, a = 7 , d = 11 – 7 = 4 and l = 139

where l = a + (n – 1)d

⇒ 139 = 7 + (n – 1)(4)

⇒ 139 = 7 + 4n – 4

⇒ 139 = 3 + 4n

⇒ 139 – 3 = 4n

⇒ 136 = 4n

Hence, the number of terms in the given AP is 34

Here, a = 41 , d = 38 – 41 = –3 and l = 8

where l = a + (n – 1)d

⇒ 8 = 41 + (n – 1)(–3)

⇒ 8 = 41 –3n + 3

⇒ 8 = 44 – 3n

⇒ 8 – 44 = –3n

⇒ –36 = –3n

Hence, the number of terms in the given AP is 12

AP = 999, 995, 991, 987,…

Here, a = 999, d = 995 – 999 = –4

a_n < 0

⇒ a + (n – 1)d < 0

⇒ 999 + (n – 1)(–4) < 0

⇒ 999 – 4n + 4 < 0

⇒ 1003 – 4n < 0

⇒ 1003 < 4n

⇒ n > 250.75

Nearest term greater than 250.75 is 251

So, 251^st term is the first negative term

Now, we will find the 251^st term

a_n = a +(n – 1)d

⇒ a₂₅₁ = 999 + (251 – 1)(–4)

⇒ a₂₅₁ = 999 + 250 × –4

⇒ a₂₅₁ = 999 – 1000

⇒ a₂₅₁ = – 1

∴, –1 is the first negative term of the given AP.

AP = 5, 8, 11, 14, …

Here, a = 5 and d = 8 – 5 = 3

Let 51 be a term, say, nth term of this AP.

We know that

a_n = a + (n – 1)d

So, 51 = 5 + (n – 1)(3)

⇒ 51 = 5 + 3n – 3

⇒ 51 = 2 + 3n

⇒ 51 – 2 = 3n

⇒ 49 = 3n

But n should be a positive integer because n is the number of terms. So, 51 is not a term of this given AP.

AP

Here, a = 4 and

Let 56 be a term, say, nth term of this AP.

We know that

a_n = a + (n – 1)d

So,

⇒ 2 × (56 – 4) = n – 1

⇒ 2 × 52 = n – 1

⇒ 104 = n – 1

⇒ 105 = n

Hence, 56 is the 105^th term of this given AP.

We have

a₇ = a + (7 – 1)d = a + 6d = 20 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = 32 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = 20 – 32

⇒ – 6d = –12

⇒ d = 2

Putting the value of d in eq (1), we get

a + 6(2) = 20

⇒ a + 12 = 20

⇒ a = 8

Hence, the required AP is 8, 10, 12, 14,…

We have

a₇ = a + (7 – 1)d = a + 6d = –4 …(1)

and a₁₃ = a + (13 – 1)d = a + 12d = –16 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 12d = –4 – (–16)

⇒ – 6d = –4 + 16

⇒ – 6d = 12

⇒ d = –2

Putting the value of d in eq (1), we get

a + 6(–2) = –4

⇒ a – 12 = –4

⇒ a = 8

Hence, the required AP is 8, 6, 4, 2,…

We have

a₈ = a + (8 – 1)d = a + 7d = 37 …(1)

and a₁₂ = a + (12 – 1)d = a + 11d = 57 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 7d – a – 11d = 37 – 57

⇒ – 4d = –20

⇒ d = 5

Putting the value of d in eq (1), we get

a + 7(5) = 37

⇒ a + 35 = 37

⇒ a = 2

Hence, the required AP is 2, 7, 12, 17,…

We have

a₇ = a + (7 – 1)d = a + 6d = 34 …(1)

and a₁₂ = a + (12 – 1)d = a + 11d = 64 …(2)

Solving the pair of linear equations (1) and (2), we get

a + 6d – a – 11d = 34 – 64

⇒ – 5d = –30

⇒ d = 6

Putting the value of d in eq (1), we get

a + 6(6) = 34

⇒ a + 36 = 34

⇒ a = –2

Hence, the required AP is –2, 4, 10, 16,…

Now, we to find the 10^th term

So, a_n = a + (n – 1)d

a₁₀ = –2 + (10 – 1)6

a₁₀ = –2 + 9 × 6

a₁₀ = 52

1^st AP = 13, 19, 25, …

Here, a = 13, d = 19 – 13 = 6

and 2^nd AP = 69, 68, 67, …

Here, a = 69, d = 68 – 69 = –1

According to the question,

13 + (n – 1)6 = 69 + (n – 1)(–1)

⇒ 13 + 6n – 6 = 69 – n + 1

⇒ 7 + 6n = 70 – n

⇒ 6n + n = 70 – 7

⇒ 7n = 63

⇒ n = 9

9^th term of the given AP’s are same.

Now, we will find the 9^th term

We have,

a_n = a + (n – 1)d

a₉ = 13 + (9 – 1)6

a₉ = 13 + 8 × 6

a₉ = 13 + 48

a₉ = 61

1^st AP = 23, 25, 27, 29, ...

Here, a = 23, d = 25 – 23 = 2

and 2^nd AP = – 17, – 10, – 3, 4, ...

Here, a = –17, d = –10 – (–17) = –10 + 17 = 7

According to the question,

23 + (n – 1)2 = –17 + (n – 1)7

⇒ 23 + 2n – 2 = –17 + 7n – 7

⇒ 21 + 2n = –24 + 7n

⇒ 2n – 7n = –24 – 21

⇒ –5n = –45

⇒ n = 9

9^th term of the given AP’s are same.

Now, we will find the 9^th term

We have,

a_n = a + (n – 1)d

a₉ = 23 + (9 – 1)2

a₉ = 23 + 8 × 2

a₉ = 23 + 16

a₉ = 39

1^st AP = 24, 20, 16, 12, ...

Here, a = 24, d = 20 – 24 = –4

and 2^nd AP = – 11, – 8, – 5, – 2, ...

Here, a = –11, d = –8 – (–11) = –8 + 11 = 3

According to the question,

24 + (n – 1)(–4) = –11 + (n – 1)3

⇒ 24 – 4n + 4 = –11 + 3n – 3

⇒ 28 – 4n = –14 + 3n

⇒ 28 + 14 = 3n + 4n

⇒ 7n = 42

⇒ n = 6

6^th term of the given AP’s are same.

Now, we will find the 6^th term

We have,

a_n = a + (n – 1)d

a₆ = 24 + (6 – 1)(–4)

a₆ = 24 + 5 × –4

a₆ = 24 – 20

a₆ = 4

1^st AP = 63, 65, 67, ...

Here, a = 63, d = 65 – 63 = 2

and 2^nd AP = 3, 10, 17, ...

Here, a = 3, d = 10 – 3 = 7

According to the question,

63 + (n – 1)2 = 3 + (n – 1)7

⇒ 63 + 2n – 2 = 3 + 7n – 7

⇒ 61 + 2n = 7n – 4

⇒ 65 = 7n – 2n

⇒ 5n = 65

⇒ n = 13

13^th term of the given AP’s are same.

Now, we will find the 13^th term

We have,

a_n = a + (n – 1)d

a₁₃ = 63 + (13 – 1)2

a₁₃ = 63 + 12 × 2

a₁₃ = 63 + 24

a₁₃ = 87

Here, a = 5 , n = 4 and

We have,

l = a + (n – 1)d

⇒19 = 10 + 6d

⇒9 = 6d

So, the missing terms are –

a₂ = a + d

a₃ = a + 2d

Hence, the missing terms are

Here, a = 54 , n = 4 and l = 42

We have,

l = a + (n – 1)d

⇒42 = 54 + (4 – 1)d

⇒42 = 54 + 3d

⇒–12 = 3d

So, the missing terms are –

a₂ = a + d = 54 – 4 = 50

a₃ = a + 2d = 54 + 2(–4) = 54 – 8 = 46

Hence, the missing terms are 50 and 46

Here, a = –4, n = 6 and l = 6

We have,

l = a + (n – 1)d

⇒6 = –4 + (6 – 1)d

⇒6 = –4 + 5d

⇒10 = 5d

So, the missing terms are –

a₂ = a + d = –4 + 2 = –2

a₃ = a + 2d = –4 + 2(2) = –4 + 4 = 0

a₄ = a + 3d = –4 + 3(2) = –4 + 6 = 2

a₅ = a + 4d = –4 + 4(2) = –4 + 8 = 4

Hence, the missing terms are –2, 0, 2 and 4

Given: a₂ = 13 and a₄ = 3

We know that,

a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

13 = a + d …(i)

and a₄ = a +(4 – 1)d

3 = a + 3d …(ii)

Solving linear equations (i) and (ii), we get

a + d – a – 3d = 13 – 3

⇒ –2d = 10

⇒ d = –5

Putting the value of d in eq. (i), we get

a – 5 = 13

⇒ a = 18

Now, a₃ = a + 2d = 18 + 2(–5) = 18 – 10 = 8

Hence, the missing terms are 18 and 8

Here, a = 7, n = 5 and l = 27

We have,

l = a + (n – 1)d

⇒27 = 7 + (5 – 1)d

⇒27 = 7 + 4d

⇒20 = 4d

So, the missing terms are –

a₂ = a + d = 7 + 5 = 12

a₃ = a + 2d = 7 + 2(5) = 7 + 10 = 17

a₄ = a + 3d = 7 + 3(5) = 7 + 15 = 22

Hence, the missing terms are 12, 17 and 22

Here, a = 2, n = 3 and l = 26

We have,

l = a + (n – 1)d

⇒26 = 2 + (3 – 1)d

⇒26 = 2 + 2d

⇒24 = 2d

So, the missing terms are –

a₂ = a + d = 2 + 12 = 14

Hence, the missing terms is 14

Given: a₃ = 13 and a₆ = 22

We know that,

a_n = a + (n – 1)d

a₃ = a + (3 – 1)d

13 = a + 2d …(i)

and a₆ = a +(6 – 1)d

22 = a + 5d …(ii)

Solving linear equations (i) and (ii), we get

a + 2d – a – 5d = 13 – 22

⇒ –3d = 9

⇒ d = 3

Putting the value of d in eq. (i), we get

a + 2(3) = 13

⇒ a + 6 = 13

⇒ a = 7

Now, a₂ = a + d = 7 + 3 = 10

a₄ = a + 3d = 7 + 3(3) = 7 + 9 = 16

a₅ = a + 4d = 7 + 4(3) = 7 + 12 = 19

Hence, the missing terms are 7, 10, 16 and 19

Here, a = –4, n = 5 and l = 6

We have,

l = a + (n – 1)d

⇒6 = –4 + (5 – 1)d

⇒6 = –4 + 4d

⇒10 = 4d

So, the missing terms are –

a₂ = a + d

a₃ = a + 2d

a₄ = a + 3d =

Hence, the missing terms are

Given: a₂ = 38 and a₆ = –22

We know that,

a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

38 = a + d …(i)

and a₆ = a +(6 – 1)d

–22 = a + 5d …(ii)

Solving linear equations (i) and (ii), we get

a + d – a – 5d = 38 – (–22)

⇒ –4d = 60

⇒ d = –15

Putting the value of d in eq. (i), we get

a + (–15) = 38

⇒ a – 15 = 38

⇒ a = 53

Now, a₃ = a + 2d = 53 + 2(–15) = 53 – 30 = 23

a₄ = a + 3d = 53 + 3(–15) = 53 – 45 = 8

a₅ = a + 4d = 53 + 4(–15) = 53 – 60 = –7

Hence, the missing terms are 53, 23, 8 and –7

Given: a₁₀ = 52 and a₁₇ = 20 + a₁₃

Now, a_n = a + (n – 1)d

a₁₀ = a + (10 – 1)d

52 = a + 9d …(i)

and a₁₇ = 20 + a₁₃

a + (17 – 1)d = 20 + a + (13 – 1)d

⇒ a + 16d = 20 + a + 12d

⇒ 16d –12d = 20

⇒ 4d = 20

⇒ d = 5

Putting the value of d in eq. (i), we get

a + 9(5) = 52

⇒ a + 45 = 52

⇒ a = 52 – 45

⇒ a = 7

Therefore, the AP is 7, 12, 17, …

Given: 3, 15, 27, 39, …

First we need to calculate 54^th term.

We know that

a_n = a + (n – 1)d

Here, a = 3, d = 15 – 3 = 12 and n = 54

So, a₅₄ = 3 + (54 – 1)12

⇒ a₅₄ = 3 + 53 × 12

⇒ a₅₄ = 3 + 636

⇒ a₅₄ = 639

Now, the term is 132 more than a₅₄ is 132 + 639 = 771

Now,

a + (n – 1)d = 771

⇒ 3 + (n – 1)12 = 771

⇒ 3 + 12n – 12 = 771

⇒ 12n = 771 + 12 – 3

⇒ 12n = 780

⇒ n = 65

Hence, the 65^th term is 132 more than the 54^th term.

Given: 3, 10, 17, 24, ...

First we need to calculate 13^th term.

We know that

a_n = a + (n – 1)d

Here, a = 3, d = 10 – 3 = 7 and n = 13

So, a₁₃ = 3 + (13 – 1)7

⇒ a₁₃ = 3 + 12 × 7

⇒ a₁₃ = 3 + 84

⇒ a₁₃ = 87

Now, the term is 84 more than a₁₃ is 84 + 87 = 171

Now,

a + (n – 1)d = 171

⇒ 3 + (n – 1)7 = 171

⇒ 3 + 7n – 7 = 171

⇒ 7n = 171 + 7 – 3

⇒ 7n = 175

⇒ n = 25

Hence, the 25^th term is 84 more than the 13^th term.

Given: a₄ = 0

To Prove: a₂₅ = 3 × a₁₁

Now, a₄ = 0

⇒ a + 3d = 0

⇒ a = –3d

We know that,

a_n = a + (n – 1)d

a₁₁ = –3d + (11 – 1)d [from (i)]

a₁₁ = –3d + 10d

a₁₁ = 7d …(ii)

Now,

a₂₅ = a + (25 – 1)d

a₂₅ = –3d + 24d [from(i)]

a₂₅ = 21d

a₂₅ = 3 × 7d

a₂₅ = 3 × a₁₁ [from(ii)]

Hence Proved

Given: 10 × a₁₀ = 15 × a₁₅

To Prove: a₂₅ = 0

Now,

10 × (a + 9d) = 15 × (a + 14d)

⇒ 10a + 90d = 15a + 210d

⇒ 10a – 15a = 210d – 90d

⇒ –5a = 120d

⇒ a = –24d …(i)

Now,

a_n = a + (n – 1)d

a₂₅ = –24d + (25 – 1)d [from (i)]

a₂₅ = –24d + 24d

a₂₅ = 0

Hence Proved

Given: a_m+1 = 2a_n+1

To Prove: a_3m+1 = 2a_m+n+1

Now,

a_n = a + (n – 1)d

⇒ a_m+1 = a + (m + 1 – 1)d

⇒ a_m+1 = a + md

and a_n+1 = a + (n + 1 – 1)d

⇒ a_n+1 = a + nd

Given: a_m+1 = 2a_n+1

a +md = 2(a + nd)

⇒ a + md = 2a + 2nd

⇒ md – 2nd = 2a – a

⇒ d(m – 2n) = a …(i)

Now,

a_m+n+1 = a + (m + n + 1 – 1)d

= a + (m + n)d

= md – 2nd + md + nd [from (i)]

= 2md – nd

a_m+n+1 = d (2m – n) …(ii)

a_3m+1 = a + (3m + 1 – 1)d

= a + 3md

= md – 2nd + 3md [from (i)]

= 4md – 2nd

= 2d( 2m – n)

a_3m+1 = 2a_m+n+1 [from (ii)]

Hence Proved

Given:

To find:

We know that,

t_n = a + (n – 1)d

So,

⇒ 3(a+3d) = 2(a+6d)

⇒ 3a + 9d = 2a + 12d

⇒ 3a – 2a = 12d – 9d

⇒ a = 3d …(i)

Now, [from (i)]

The list of 3 digit numbers divisible by 5 is:

100, 105, 110,…,995

Here a = 100, d = 105 – 100 = 5, a_n = 995

We know that

a_n = a + (n – 1)d

995 = 100 + (n – 1)5

⇒ 895 = (n – 1)5

⇒ 179 = n – 1

⇒ 180 = n

So, there are 180 three– digit numbers divisible by 5.

The list of 3 digit numbers divisible by 7 is:

105, 112, 119,…,994

Here a = 105, d = 112 – 105 = 7, a_n = 994

We know that

a_n = a + (n – 1)d

994 = 105 + (n – 1)7

⇒ 889 = (n – 1)7

⇒ 127 = n – 1

⇒ 128 = n

So, there are 128 three– digit numbers divisible by 7.

To show: t_m + t_2n+m = 2 t_m+n

Taking LHS

t_m + t_2n+m = a + (m – 1)d + a + (2n + m – 1)d

= 2a + md – d + 2nd + md – d

= 2a + 2md + 2nd – 2d

= 2 {a + (m + n – 1)d}

= 2t_m+n

= RHS

∴LHS = RHS

Hence Proved

Let 5a + 2, 4a – 1, a + 2 are in AP

So, first term a = 5a + 2

d = 4a – 1 – 5a – 2 = – a – 3

n = 3

l = a + 2

So,

l = a + (n – 1)d

⇒ a + 2 =5a + 2 + (3 – 1)(–a – 3)

⇒ a + 2 – 5a – 2 = –3a – 9 + a + 3

⇒ – 4a = –2a – 6

⇒ – 4a + 2a = – 6

⇒ –2a = – 6

⇒ a = 3

We know that nth term of an A.P is given by,

a_n = a + (n – 1) d

Now equating it with the expression given we get,

2 n + 1 = a + (n – 1) d

2 n + 1 = a + nd – d

2 n + 1 = nd + (a – d)

Equating both sides we get,

d = 2 and a – d = 1

So we get,

a = 3 and d = 2.

So the first term of this sequence is 3, and the common difference is 2.

Given: a₄ + a₈ = 24

⇒ a +3d + a + 7d = 24

⇒ 2a + 10d = 24 …(i)

and a₆ + a₁₀ = 44

⇒ a +5d + a + 9d = 44

⇒ 2a + 14d = 44 …(ii)

Solving Linear equations (i) and (ii), we get

2a + 10d – 2a – 14d = 24 – 44

⇒ –4d = – 20

⇒ d = 5

Putting the value of d in eq. (i), we get

2a + 10×5 = 24

⇒ 2a + 50 = 24

⇒ 2a =24 –50

⇒ 2a =–26

⇒ a = –13

So, the first three terms are –13, –8, –3.

Let the required number of years = n

Given t_n = 1500, a= 700, d = 40

We know that,

t_n = a +(n – 1)d

⇒ 1500 = 700 + (n – 1)40

⇒ 800 = (n – 1)40

⇒ 20 = n – 1

⇒ n = 21

Hence, in 21years he will reach maximum of the scale.

Let the required sum = a

and the interest carried every year = d

According to question,

In 4years, a sum of money kept in bank account = Rs. 600

i.e. t₅ = 600 ⇒ a + 4d = 600 …(i)

and in 12 years , sum of money kept = Rs. 800

i.e. t₁₃ = 800 ⇒ a + 12d = 800 …(ii)

Solving linear equations (i) and (ii), we get

a + 4d – a – 12d = 600 – 800

⇒ – 8d = –200

⇒ d = 25

Putting the value of d in eq.(i), we get

a + 4(25) = 600

⇒ a + 100 = 600

⇒ a = 500

Hence, the sum and interest carried every year is Rs 500 and Rs 25 respectively.

The first instalment of the loan = Rs. 100

The 2^nd instalment of the loan = Rs. 105

The 3^rd instalment of the loan = Rs. 110

and so, on

The amount that the man repays every month forms an AP.

Therefore, the series is

100, 105, 110, 115,…

Here, a = 100, d = 105 – 100 = 5

We know that,

a_n = a + (n – 1)d

a₃₀ = 100 + (30 – 1)5

⇒ a₃₀ = 100 + 29 × 5

⇒ a₃₀ = 100 +145

⇒ a₃₀ = 245

Hence, the amount he will pay in the 30^th installment is Rs 245.

Let the three numbers are in AP = a, a + d, a + 2d

According to the question,

The sum of three terms = 27

⇒ a + (a + d) + (a + 2d) = 27

⇒ 3a + 3d = 27

⇒ a + d = 9

⇒ a = 9 – d …(i)

and the sum of their squares = 275

⇒ a² + (a + d)² + (a + 2d)² = 275

⇒ (9 – d)² + (9)² + ( 9 – d + 2d)² = 275 [from(i)]

⇒ 81 + d² – 18d + 81 + 81 + d² + 18d = 275

⇒ 243 + 2d² = 275

⇒ 2d² = 275 – 243

⇒ 2d² = 32

⇒ d² = 16

⇒ d = √16

⇒ d = ±4

Now, if d = 4, then a = 9 – 4 = 5

and if d = – 4, then a = 9 – ( – 4) = 9 + 4 = 13

So, the numbers are →

if a = 5 and d = 4

5, 9, 13

and if a = 13 and d = – 4

13, 9, 5

Let the three numbers are in AP = a, a + d, a + 2d

According to the question,

The sum of three terms = 12

⇒ a + (a + d) + (a + 2d) = 12

⇒ 3a + 3d = 12

⇒ a + d = 4

⇒ a = 4 – d …(i)

and the sum of their cubes = 408

⇒ a³ + (a + d)³ + (a + 2d)³ = 408

⇒ (4 – d)³ + (4)³ + ( 4 – d + 2d)³ = 408 [from(i)]

⇒ (4 – d)³ + (4)³ + ( 4 + d)³ = 408

⇒ 64 – d³ + 12d² – 48d + 64 + 64 + d³ + 12d² + 48d = 408

⇒ 192 + 24d² = 408

⇒ 24d² = 408 – 192

⇒ 24d² = 216

⇒ d² = 9

⇒ d = √9

⇒ d = ±3

Now, if d = 3, then a = 4 – 3 = 1

and if d = – 3, then a = 4 – ( – 3) = 4 + 3 = 7

So, the numbers are →

if a = 1 and d = 3

1, 4, 7

and if a = 7 and d = – 3

7, 4, 1

Let the middle term = a and the common difference = d

The first term = a – d and the succeeding term = a + d

So, the three parts are a – d, a, a + d

According to the question,

Sum of these three parts = 15

⇒ a – d + a + a + d = 15

⇒ 3a = 15

⇒ a = 5

and the sum of their squares = 83

⇒ (a – d)² + a² + (a + d)² = 83

⇒ (5 – d)² + (5)² + ( 5 + d)² = 83 [from(i)]

⇒ 25 + d² – 10d + 25 + 25 + d² + 10d = 83

⇒ 75 + 2d² = 83

⇒ 2d² = 83 – 75

⇒ 2d² = 8

⇒ d² = 4

⇒ d = √4

⇒ d = ±2

Case: (i) If d = 2, then

a – d = 5 – 2 = 3

a = 5

a + d = 5 + 2 = 7

Hence, the three parts are

3, 5, 7

Case: (ii) If d = – 2, then

a – d = 5 – ( – 2) = 7

a = 5

a + d = 5 + ( – 2) = 3

Hence, the three parts are

7, 5, 3

Let the four parts which are in AP are

(a – 3d), (a – d), (a + d), (a + 3d)

According to question,

The sum of these four parts = 20

⇒(a – 3d) + (a – d) + (a + d) + (a + 3d) = 20

⇒ 4a = 20

⇒ a = 5 …(i)

Now, it is also given that

product of the first and fourth : product of the second and third = 2 : 3

i.e. (a – 3d) × (a + 3d) : (a – d) × (a + d) = 2 : 3

[∵(a – b)(a + b) = a² – b² ]

⇒ 3(a² – 9d²) = 2(a² – d²)

⇒ 3a² – 27d² = 2a² – 2d²

⇒ 3a² – 2a² = – 2d² + 27d²

⇒ (5)² = – 2d² + 27d² [from (i)]

⇒ 25 = 25d²

⇒ 1 = d²

⇒ d = ±1

Case I: if d = 1 and a = 5

a – 3d = 5 – 3(1) = 5 – 3 = 2

a – d = 5 – 1 = 4

a + d = 5 + 1 = 6

a + 3d = 5 + 3(1) = 5 + 3 = 8

Hence, the four parts are

2, 4, 6, 8

Case II: if d = – 1 and a = 5

a – 3d = 5 – 3( – 1) = 5 + 3 = 8

a – d = 5 – ( – 1) = 5 + 1 = 6

a + d = 5 + ( – 1) = 5 – 1 = 4

a + 3d = 5 + 3( – 1) = 5 – 3 = 2

Hence, the four parts are

8, 4, 6, 2

Let the three numbers are (a – d), a and (a + d)

According to question,

Sum of these three numbers = 21

⇒ a – d + a + a + d = 21

⇒ 3a = 21

⇒ a = 7 …(i)

and it is also given that

Product of these numbers = 231

⇒(a – d) × a × (a + d) = 231

⇒(7 – d) × 7 × (7 + d) = 231

⇒ 7 × (7² – d²) = 231 [∵ (a – b)(a + b) = a² – b²]

⇒ 7 × (49 – d²) = 231

⇒ 343 – 7d² = 231

⇒ – 7d² = 231 – 343

⇒ – 7d² = – 112

⇒ d² = 16

⇒ d = √16

⇒ d = ±4

Case I: If d = 4 and a = 7

a – d = 7 – 4 = 3

a = 7

a + d = 7 + 4 = 11

So, the numbers are

3, 7, 11

Case II: If d = – 4 and a = 7

a – d = 7 – ( – 4) = 7 + 4 = 11

a = 7

a + d = 7 + ( – 4) = 7 – 4 = 3

So, the numbers are

11, 7, 3

Let the three numbers are (a – d), a and (a + d)

According to question,

Sum of these three numbers = 3

⇒ a – d + a + a + d = 3

⇒ 3a = 3

⇒ a = 1 …(i)

and it is also given that

Product of these numbers = – 35

⇒(a – d) × a × (a + d) = – 35

⇒(1 – d) × 1 × (1 + d) = – 35

⇒ 1 × (1² – d²) = – 35 [∵ (a – b)(a + b) = a² – b²]

⇒ 1 × (1 – d²) = – 35

⇒ 1 – d² = – 35

⇒ – d² = – 35 – 1

⇒ – d² = – 36

⇒ d² = 36

⇒ d = √36

⇒ d = ±6

Case I: If d = 6 and a = 1

a – d = 1 – 6 = – 5

a = 1

a + d = 1 + 6 = 7

So, the numbers are

– 5, 1, 7

Case II: If d = – 6 and a = 1

a – d = 1 – ( – 6) = 1 + 6 = 7

a = 1

a + d = 1 + ( – 6) = 1 – 6 = – 5

So, the numbers are

7, 1, – 5

Given: a + b + c ≠ 0

and are in AP

To Prove: are in AP

if are in AP

[multiplying each term by a + b + c]

i.e. if are in AP

which is given to be true

Hence, are in AP

a², b², c² are in AP

∴ b² – a² = c² – b²

⇒(b – a)(b + a) = (c – b)(c + b)

Hence Proved

Given: a, b, c are in AP

∴ b – a = c – b …(i)

To Prove: are in AP

⇒ b – a = c – b

∴a, b, c are in AP

are in AP

Given: a, b, c are in AP

Since, a, b, c are in AP, we have a + c = 2b …(i)

Now, (b + c)² – a², (c + a)² – b², (a + b)² – c² will be in A.P

If (b + c – a)(b + c + a), (c + a – b)(c + a + b), (a + b – c)(a + b + c) are in AP

i.e. if b + c – a, c + a – b, a + b – c are in AP

[dividing by (a + b + c)]

if (b + c – a) + (a + b – c) = 2(c + a – b)

if 2b = 2(c + a – b)

if b = c + a – b

if a + c = 2b which is true by (i)

Hence, (b + c)² – a², (c + a)² – b², (a + b)² – c² are in A.P

Given: a, b, c are in AP

Since, a, b, c are in AP, we have a + c = 2b …(i)

To Prove : are in AP

⇒

⇒ (√b – √a)(√b + √a) = (√c – √b)(√c + √b)

⇒ b – a = c – b

⇒ 2b = a + c, which is True ... from (i)

Hence, the result.

Given: are in AP

Taking LCM

⇒b²c + c²b + a²b + ab² – 2ac² – 2a²c = 0

⇒ b²c + c²b + a²b + ab² –ac² – ac² – a²c – a²c = 0

⇒(b²c – a²c) + (c²b – ac²) + (a²b – a²c) + (ab² – ac²) = 0

⇒ c (b – a)(b + a) + c²(b – a) + a² (b – c) + a(b + c)(b – c) = 0

⇒ c(b – a) {(b + a) + c} + a(b – c) {a + (b + c)} = 0

⇒ (a + b + c){cb – ca + ab – ca} = 0

Given a + b + c ≠ 0

⇒cb – ca + ab – ca = 0

⇒cb – 2ca + ab = 0

are in AP

Hence Proved

Given: (b – c)², (c – a)², (a – b)² are in A.P

∴ 2(c – a)² = (b – c)² + (a – b)² …(i)

To Prove: are in AP

or

⇒2(b – c)(a – b) = (a – c)(c – a)

⇒2[ab – b² – ca + cb] = ac – a² – c² + ac

⇒2ab – 2b² – 2ac + 2cb = 2ac – a² – c²

⇒ a² + c² – 4ac = 2b² – 2ab – 2cb

Adding both sides, a² + c², we get

⇒2(a² + c²) – 4ac = a² + b² – 2ab + c² + b²– 2cb

⇒ 2 (a – c)² = ( b – a)² + (b – c)² which is true from (i)

∴(b – c)², (c – a)², (a – b)² are in A.P

are in AP

Hence Proved

Given: a, b, c are in AP

∴ a + c = 2b

…(i)

Now taking RHS i.e. 4(a – b)(b – c)

[from(i)]

⇒(a – c)²

= LHS

Hence Proved

Given: a, b, c are in AP

∴ a + c = 2b …(i)

…(ii)

Taking Lhs i.e. a³ + c³ + 6abc

[from (i)]

⇒ a³ + c³ + 3ac(a + c)

⇒ a³ + c³ 3a²c + 3ac²

⇒ (a + c)³

⇒ (2b)³ [from (ii)]

= 8b³ = RHS

Hence Proved

Given: a, b, c are in AP

∴ a + c = 2b …(i)

…(ii)

Now, taking LHS i.e. (a + 2b — c)(2b + c — a)(c + a — b)

[from (ii)]

⇒ 4abc

[from (ii)]

= RHS

Hence Proved

Taking n = 1, we get

⇒ S₁ = 4

⇒ a₁ = 4

Taking n = 2, we get

⇒ S₂ = 13

∴ a₂ = S₂ – S₁ = 13 – 4 = 9

Taking n = 3, we get

⇒ S₃ = 27

∴ a₃ = S₃ – S₂ = 27 – 13 = 14

So, a = 4,

d = a₂ – a₁ = 9 – 4 = 5

Now, we have to find the 20^th term

a_n = a + (n – 1)d

a₂₀ = 4 + (20 – 1)5

a₂₀ = 4 + 19 × 5

a₂₀ = 4 + 95

a₂₀ = 99

Hence, the 20^th term is 99.

S_n = 3n² + 2n

Taking n = 1, we get

S₁ = 3(1)² + 2(1)

⇒ S₁ = 3 + 2

⇒ S₁ = 5

⇒ a₁ = 5

Taking n = 2, we get

S₂ = 3(2)² + 2(2)

⇒ S₂ = 12 + 4

⇒ S₂ = 16

∴ a₂ = S₂ – S₁ = 16 – 5 = 11

Taking n = 3, we get

S₃ = 3(3)² + 2(3)

⇒ S₃ = 27 + 6

⇒ S₃ = 33

∴ a₃ = S₃ – S₂ = 33 – 16 = 17

So, a = 5,

d = a₂ – a₁ = 11 – 5 = 6

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a₁₅ = 5 + (15 – 1)6

a₁₅ = 5 + 14 × 6

a₁₅ = 5 + 84

a₁₅ = 89

Hence, the 15^th term is 89 and AP is 5, 11, 17, 23,…

S_n =2n² + 5n

Taking n = 1, we get

S₁ = 2(1)² + 5(1)

⇒ S₁ = 2 + 5

⇒ S₁ = 7

⇒ a₁ = 7

Taking n = 2, we get

S₂ = 2(2)² + 5(2)

⇒ S₂ = 8 + 10

⇒ S₂ = 18

∴ a₂ = S₂ – S₁ = 18 – 7 = 11

Taking n = 3, we get

S₃ = 2(3)² + 5(3)

⇒ S₃ = 18 + 15

⇒ S₃ = 33

∴ a₃ = S₃ – S₂ = 33 – 18 = 15

So, a = 7,

d = a₂ – a₁ = 11 – 7 = 4

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a_n = 7 + (n – 1)4

a_n = 7 + 4n – 4

a_n = 3 + 4n

Hence, the n^th term is 4n + 3.

S_n = 3n² + 5n

Taking n = 1, we get

S₁ = 3(1)² + 5(1)

⇒ S₁ = 3 + 5

⇒ S₁ = 8

⇒ a₁ = 8

Taking n = 2, we get

S₂ = 3(2)² + 5(2)

⇒ S₂ = 12 + 10

⇒ S₂ = 22

∴ a₂ = S₂ – S₁ = 22 – 8 = 14

Taking n = 3, we get

S₃ = 3(3)² + 5(3)

⇒ S₃ = 27 + 15

⇒ S₃ = 42

∴ a₃ = S₃ – S₂ = 42 – 22 = 20

So, a = 8,

d = a₂ – a₁ = 14 – 8 = 6

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a₁₆ = 8 + (16 – 1)6

a₁₆ = 8 + 15 × 6

a₁₆ = 8 + 90

a₁₆ = 98

Hence, the 16^th term is 98.

S_n = 3n² – n

Taking n = 1, we get

S₁ = 3(1)² - (1)

⇒ S₁ = 3 – 1

⇒ S₁ = 2

⇒ a₁ = 2

Taking n = 2, we get

S₂ = 3(2)² – 2

⇒ S₂ = 12 – 2

⇒ S₂ = 10

∴ a₂ = S₂ – S₁ = 10 – 2 = 8

Taking n = 3, we get

S₃ = 3(3)² – 3

⇒ S₃ = 27 – 3

⇒ S₃ = 24

∴ a₃ = 24 – 10 = 14

So, a = 1,

d = a₂ – a₁ = 8 - 2 = 6

Now, we have to find the 15^th term

a_n = a + (n – 1)d

a_n = 2 + (n – 1)6

a_n = 2 + 6n – 6

a_n = - 4 + 6n

Hence, the n^th term is 4n - 3.

Taking n = 1, we get

⇒ S₁ = 4

⇒ a₁ = 4

Taking n = 2, we get

⇒ S₂ = 11

∴ a₂ = S₂ – S₁ = 11 – 4 = 7

Taking n = 3, we get

⇒ S₃ = 21

∴ a₃ = S₃ – S₂ = 21 – 11 = 10

So, a = 4,

d = a₂ – a₁ = 7 – 4 = 3

Now, we have to find the 25^th term

a_n = a + (n – 1)d

a₂₅ = 4 + (25 – 1)3

a₂₅ = 4 + 24 × 3

a₂₅ = 4 + 72

a₂₅ = 76

Hence, the 25^th term is 76.

Given: a_n = 2n + 1

Taking n = 1,

a₁ = 2(1) + 1 = 2 + 1 = 3

Taking n = 2,

a₂ = 2(2) + 1 = 4 + 1 = 5

Taking n = 3,

a₃ = 2(3) + 1 = 6 + 1 = 7

Therefore the series is 3, 5, 7, …

So, a = 3, d = a₂ – a₁ = 5 – 3 = 2

Now, we have to find the sum of first n terms of the AP

⇒ S_n = 2n + n²

Hence, the sum of n terms is n² + 2n.

Given: a_n = 9 – 5n

Taking n = 1,

a₁ = 9 – 5(1) = 9 – 5 = 4

Taking n = 2,

a₂ = 9 – 5(2) = 9 – 10 = -1

Taking n = 3,

a₃ = 9 – 5(3) = 9 – 15 = -6

Therefore the series is 4, -1, -6, …

So, a = 4, d = a₂ – a₁ = -1 – 4 = -5

Now, we have to find the sum of the first 15 terms of the AP

⇒ S₁₅ = 15 × (-31)

⇒ S₁₅ = -465

Hence, the sum of 15 terms is -465.

Given: a_n = 1 – 4n

Taking n = 1,

a₁ = 1 – 4(1) = 1 – 4 = -3

Taking n = 2,

a₂ = 1 – 4(2) = 1 – 8 = -7

Taking n = 3,

a₃ = 1 – 4(3) = 1 – 12 = -11

Therefore the series is -3, -7, -11, …

So, a = -3, d = a₂ – a₁ = -7 – (-3) = -7 + 3 = -4

Now, we have to find the sum of the first 25 terms of the AP

⇒ S₂₅ = 25 × (-51)

⇒ S₂₅ = -1275

Hence, the sum of 25 terms is -1275.

Given: S_n = n² + 2n …(i)

S_n-1 = (n – 1)² + 2(n – 1) = n² + 1 – 2n + 2n – 2 = n² - 1 …(ii)

Subtracting eq (ii) from (i), we get

t_n = S_n – S_n-1 = n² + 2n – n² + 1 = 2n + 1

The nth term of an AP is 2n + 1.

As it is given that kth term of the AP = 5k + 1

∴ a_k = a + (k – 1)d

⇒ 5k + 1 = a + (k – 1)d

⇒ 5k + 1 = a + kd – d

Now, on comparing the coefficient of k, we get

d = 5

and a – d = 1

⇒ a – 5 = 1

⇒ a = 6

We know that,

S_n = 3n² + 5n

Taking n = 1, we get

S₁ = 3(1)² + 5(1)

⇒ S₁ = 3 + 5

⇒ S₁ = 8

⇒ a₁ = 8

Taking n = 2, we get

S₂ = 3(2)² + 5(2)

⇒ S₂ = 12 + 10

⇒ S₂ = 22

∴ a₂ = S₂ – S₁ = 22 – 8 = 14

Taking n = 3, we get

S₃ = 3(3)² + 5(3)

⇒ S₃ = 27 + 15

⇒ S₃ = 42

∴ a₃ = S₃ – S₂ = 42 – 22 = 20

So, a = 8,

d = a₂ – a₁ = 14 – 8 = 6

Now, we have to find the value of m

a_n = a + (n – 1)d

⇒ a_m = 8 + (m – 1)6

⇒ 164 = 8 + 6m – 6

⇒ 164 = 2 + 6m

⇒ 162 = 6m

⇒ m = 27

S_n = qn² + pn

Taking n = 1, we get

S₁ = q(1)² + p(1)

⇒ S₁ = q + p

⇒ a₁ = q + p

Taking n = 2, we get

S₂ = q(2)² + p(2)

⇒ S₂ = 4q + 2p

∴ a₂ = S₂ – S₁ = 4q + 2p – q - p = 3q + p

Taking n = 3, we get

S₃ = q(3)² + p(3)

⇒ S₃ = 9q + 3p

∴ a₃ = S₃ – S₂ = 9q + 3p – 4q – 2p = 5q + p

So, a = q + p,

d = a₂ – a₁ = 3q + p – (q + p) = 3q + p – q – p = 2q

Hence, the common difference is 2q.

Taking n = 1, we get

⇒ S₁ = P

⇒ a₁ = P

Taking n = 2, we get

⇒ S₂ = 2P + Q

∴ a₂ = S₂ – S₁ = 2P + Q – P = P + Q

Taking n = 3, we get

⇒ S₃ = 3P + 3Q

∴ a₃ = S₃ – S₂ = 3P + 3Q – 2P – Q = P + 2Q

So, a = P,

d = a₂ – a₁ = P + Q – (P) = Q

= a₃ – a₂ = P + 2Q – (P + Q) = P + 2Q – P – Q = Q

Hence, the common difference is Q.

Here, a = 25, d = 28 – 25 = 3 and a_n = 100

We know that,

a_n = a + (n – 1)d

⇒ 100 = 25 + (n – 1)3

⇒ 75 = (n – 1)3

⇒ 25 = n – 1

⇒ 26 = n

Now,

⇒ S₂₆ = 13[50 + 25 × 3]

⇒ S₂₆ = 13[50 + 75]

⇒ S₂₆ = 13 × 125

⇒ S₂₆ = 1625

Let a_n = 89

AP = 4, 9, 14, …89

Here, a = 4, d = 14 – 9 = 5

We know that

a_n = a + (n – 1)d

⇒ 89 = 4 + (n – 1)5

⇒ 85 = (n – 1)5

⇒ 17 = n – 1

⇒ 18 = n

So, 89 is the 18^th term of the given AP

Now, we find the sum of 4 + 9 + 14 + … + 89

We know that,

⇒ S₁₈ = 9[8 + 17 × 5]

⇒ S₁₈ = 9[8 + 85]

⇒ S₁₈ = 9 × 93

⇒ S₁₈ = 837

Hence, the sum of the given AP is 837.

Here, a = 1, d = 6 – 1 = 5 and S_n = 148

⇒ 296 = n[5n – 3]

⇒ 5n² – 3n – 296 = 0

⇒ 5n² – 40n + 37n – 296 = 0

⇒ 5n(n – 8) + 37(n – 8) = 0

⇒ (5n + 37)(n – 8) = 0

⇒ 5n + 37 = 0 or n – 8 = 0

⇒ or n = 8

But is not a positive integer.

∴ n = 8

⇒ x = a₈ = a + 7d = 1 + 7 × 5 = 1 + 35 = 36

Hence, x = 36

Here, a = 25, d = 22 – 25 = -3 and S_n = 115

⇒ 230 = n[53 – 3n]

⇒ 3n² – 53n + 230 = 0

⇒ 3n² – 30n - 23n + 230 = 0

⇒ 3n(n – 10) - 23(n – 10) = 0

⇒ (3n – 23)(n – 10) = 0

⇒ 3n – 23 = 0 or n – 10 = 0

⇒ or n = 10

But is not an integer.

∴ n = 10

⇒ x = a₁₀ = a + 9d = 25 + 9 × (-3) = 25 – 27 = -2

Hence, x = -2

AP = 64, 60, 56, …

Here, a = 64, d = 60 – 64 = -4

⇒ 1088 = n[132 – 4n]

⇒ 4n² – 132n + 1088 = 0

⇒ n² – 33n + 272= 0

⇒ n² – 16n - 17n + 272 = 0

⇒ n(n – 16) - 17(n – 16) = 0

⇒ (n – 16)(n – 17) = 0

⇒ n – 16 = 0 or n – 17 = 0

⇒ n = 16 or n = 17

If n = 16, a = 64 and d = -4

a₁₆ = 64 + (16 – 1)(-4)

a₁₆ = 64 + 15 × -4

a₁₆ = 64 – 60

a₁₆ = 4

and If n = 17, a = 64 and d = -4

a₁₇ = 64 + (17 – 1)(-4)

a₁₇ = 64 + 16 × -4

a₁₇ = 64 – 64

a₁₇ = 0

Now, we will check at which term the sum of the AP is 544.

⇒ S₁₆ = 8[68]

⇒ S₁₆ = 544

and

⇒ S₁₇ = 17 × 32

⇒ S₁₇ = 544

So, the terms may be either 17 or 16 both holds true.

We get a double answer because the 17^th term is zero and when we add this in the sum, the sum remains the same.

AP = 3, 5, 7, 9, …

Here, a = 3, d = 5 – 3 = 2 and S_n = 120

We know that,

⇒ 120 = n[2+n]

⇒ n² + 2n – 120 = 0

⇒ n² + 12n – 10n – 120 = 0

⇒ n(n + 12) - 10(n + 12) = 0

⇒ (n – 10)(n + 12) = 0

⇒ n – 10 = 0 or n + 12 = 0

⇒ n = 10 or n = -12

But number of terms can’t be negative. So, n = 10

Hence, for n = 10 the sum is 120 for the given AP.

AP = 63, 60, 57,…

Here, a = 63, d = 60 – 63 = -3 and S_n = 693

We know that,

⇒ 1386 = n[129 – 3n]

⇒ 3n² – 129n + 1386 = 0

⇒ n² – 43n + 462 = 0

⇒ n² – 22n – 21n + 462 = 0

⇒ n(n – 22) - 21(n – 22) = 0

⇒ (n – 21)(n – 22) = 0

⇒ n – 21 = 0 or n – 22 = 0

⇒ n = 21 or n = 22

So, n = 21 and 22

If n = 21, a = 63 and d = -3

a₂₁ = 63 + (21 – 1)(-3)

a₂₁ = 63 + 20 × -3

a₂₁ = 63 – 60

a₂₁ = 3

and If n = 22, a = 63 and d = -3

a₂₂ = 63 + (22 – 1)(-3)

a₂₂ = 63 + 21 × -3

a₂₂ = 63 – 63

a₂₂ = 0

Now, we will check at which term the sum of the AP is 693.

⇒ S₂₁ = 21 × 33

⇒ S₂₁ = 693

and

⇒ S₂₂ = 11 × 63

⇒ S₂₂ = 693

So, the terms may be either 21 or 22 both holds true.

We get the double answer because here the 22^nd term is zero and it does not affect the sum.

Here, a = 15, d = 12 – 15 = -3 and S_n = 15

We know that,

⇒ 30 = n[33 – 3n ]

⇒ 3n² – 33n + 30 = 0

⇒ 3n² – 30n – 3n + 30 = 0

⇒ 3n(n – 10) -3(n – 10) = 0

⇒ (n – 10)(3n – 3) = 0

⇒ n – 10 = 0 or 3n – 3 = 0

⇒ n = 10 or n = 1

The number of terms can be 1 or 10.

Here, the common difference is negative.

∴ The AP starts from a positive term, and its terms are decreasing.

∴ All the terms after 6^th term are negative.

We get a double answer because these positive terms from 2^nd to 5^th term when added to negative terms from 7^th to 10^th term, they cancel out each other and the sum remains same.

The odd numbers lying between 100 and 200 are

101, 103, 105,…, 199

a₂ – a₁ = 103 – 101 = 2

a₃ – a₂ = 105 – 103 = 2

∵ a₃ – a₂ = a₂ – a₁ = 2

Therefore, the series is in AP

Here, a = 101, d = 2 and a_n = 199

We know that,

a_n = a + (n – 1)d

⇒ 199 = 101 + (n – 1)2

⇒ 199 – 101 = (n – 1)2

⇒ 98 = (n – 1)2

⇒ 49 = (n – 1)

⇒ n = 50

Now, we have to find the sum of this AP

⇒ S₅₀ = 25[202 + 49 × 2]

⇒ S₅₀ = 25[300]

⇒ S₅₀ = 7500

Hence, the sum of all odd numbers lying between 100 and 200 is 7500.

The odd numbers lying between 1 and 2001 are

1, 3, 5,…, 2001

a₂ – a₁ = 3 – 1 = 2

a₃ – a₂ = 5 – 3 = 2

∵ a₃ – a₂ = a₂ – a₁ = 2

Therefore, the series is in AP

Here, a = 1, d = 2 and a_n = 2001

We know that,

a_n = a + (n – 1)d

⇒ 2001 = 1 + (n – 1)2

⇒ 2001 – 1 = (n – 1)2

⇒ 2000 = (n – 1)2

⇒ 1000 = (n – 1)

⇒ n = 1001

Now, we have to find the sum of this AP

⇒ S₁₀₀₁ = 1001[1 + 1000]

⇒ S₁₀₀₁ = 1001 [1001]

⇒ S₁₀₀₁ = 1002001

Hence, the sum of all odd numbers lying between 1 and 2001 is 1002001.

Given: a₂ = 2 and a₇ = 22 and n = 35

We know that,

a₂ = a + d = 2 …(i)

and a₇ = a + 6d = 22 …(ii)

Solving the linear equations (i) and (ii), we get

a + d – a – 6d = 2 – 22

⇒ - 5d = -20

⇒ d =4

Putting the value of d in eq. (i), we get

a + 4 = 2

⇒ a = 2 – 4 = -2

Now, we have to find the sum of first 35 terms.

⇒ S₃₅ = 35 [-2 + 34 × 2]

⇒ S₃₅ = 35 [66]

⇒ S₃₅ = 2310

Given: S_p = q and S_q = p

To find: S_p+q

We know that,

…(i)

Now,

…(ii)

From eq. (i) and (ii), we get

[∵, a² – b² = (a – b)(a + b)]

…(iii)

Now, putting the value of d in eq. (i), we get

…(iv)

Now, we to find S_p+q

[from (iii) & (iv)]

⇒ S_p+q = - (p+q)

Hence, the sum of first (p+q) terms is –(p + q)

Here, a = -6,

and S_n = -25

We know that,

⇒ -100 = n[-25 + n]

⇒ n² – 25n + 100 = 0

⇒ n² – 20n – 5n + 100 = 0

⇒ n(n – 20) - 5(n – 20) = 0

⇒ (n – 20)(n – 5) = 0

⇒ n – 5 = 0 or n – 20 = 0

⇒ n = 5 or n = 20

So, n = 5 or 20

The numbers lying between 107 and 253 that are multiples of 5 are

110, 115, 120,…, 250

a₂ – a₁ = 115 – 110 = 5

a₃ – a₂ = 120 – 115 = 5

∵ a₃ – a₂ = a₂ – a₁ = 5

Therefore, the series is in AP

Here, a = 110, d = 5 and a_n = 250

We know that,

a_n = a + (n – 1)d

⇒ 250 = 110 + (n – 1)5

⇒ 250 – 110 = (n – 1)5

⇒ 140 = (n – 1)5

⇒ 28 = (n – 1)

⇒ n = 29

Now, we have to find the sum of this AP

⇒ S₂₉ = 29[110 + 14 × 5]

⇒ S₂₉ = 29[180]

⇒ S₂₉ = 5220

Hence, the sum of all numbers lying between 107 and 253 is 5220.

The numbers lying between 100 and 1000 that are multiples of 5 are

105, 110, 115, 120,…, 995

a₂ – a₁ = 110 – 105 = 5

a₃ – a₂ = 115 – 110 = 5

∵ a₃ – a₂ = a₂ – a₁ = 5

Therefore, the series is in AP

Here, a = 105, d = 5 and a_n = 995

We know that,

a_n = a + (n – 1)d

⇒ 995 = 105 + (n – 1)5

⇒ 995 – 105 = (n – 1)5

⇒ 890 = (n – 1)5

⇒ 178 = (n – 1)

⇒ n = 179

Now, we have to find the sum of this AP

⇒ S₁₇₉ = 179[105 + 89 × 5]

⇒ S₁₇₉ = 179 [550]

⇒ S₁₇₉ = 98450

Hence, the sum of all numbers lying between 100 and 1000 that are multiples of 5 is 98450.

The two digit odd positive integers are

11, 13, 15,…, 99

a₂ – a₁ = 13 – 11 = 2

a₃ – a₂ = 15 – 13 = 2

∵ a₃ – a₂ = a₂ – a₁ = 2

Therefore, the series is in AP

Here, a = 11, d = 2 and a_n = 99

We know that,

a_n = a + (n – 1)d

⇒ 99 = 11 + (n – 1)2

⇒ 99 – 11 = (n – 1)2

⇒ 88 = (n – 1)2

⇒ 44 = (n – 1)

⇒ n = 45

Now, we have to find the sum of this AP

⇒ S₄₅ = 45[11 + 44]

⇒ S₄₅ = 45[55]

⇒ S₄₅ = 2475

Hence, the sum of all two digit odd numbers are 2475.

The numbers lying between 300 and 700 which are multiples of 9 are

306, 315, 324,…, 693

a₂ – a₁ = 315 – 306 = 9

a₃ – a₂ = 324 – 315 = 9

∵ a₃ – a₂ = a₂ – a₁ = 9

Therefore, the series is in AP

Here, a = 306, d = 9 and a_n = 693

We know that,

a_n = a + (n – 1)d

⇒ 693 = 306 + (n – 1)9

⇒ 693 - 306 = (n – 1)9

⇒ 387 = (n – 1)9

⇒ 43 = (n – 1)

⇒ n = 44

Now, we have to find the sum of this AP

⇒ S₄₄ = 22[612 + 387]

⇒ S₄₄ = 22[999]

⇒ S₄₄ = 21978

Hence, the sum of all numbers lying between 300 and 700 is 21978.

The three digit natural numbers which are multiples of 7 are

105, 112, 119,…, 994

a₂ – a₁ = 112 – 105 = 7

a₃ – a₂ = 112 – 105 = 7

∵ a₃ – a₂ = a₂ – a₁ = 7

Therefore, the series is in AP

Here, a = 105, d = 7 and a_n = 994

We know that,

a_n = a + (n – 1)d

⇒ 994 = 105 + (n – 1)7

⇒ 994 – 105 = (n – 1)7

⇒ 889 = (n – 1)7

⇒ 127 = (n – 1)

⇒ n = 128

Now, we have to find the sum of this AP

⇒ S₁₂₈ = 64[210 + 127 × 7]

⇒ S₁₂₈ = 64[1099]

⇒ S₁₂₈ = 70336

Hence, the sum of all three digit numbers which are multiples of 7 are 70336.

The numbers lying between 100 and 500 which are divisible by 8 are

104, 112, 120, 128, 136,…, 496

a₂ – a₁ = 112 – 104 = 8

a₃ – a₂ = 120 – 112 = 8

∵ a₃ – a₂ = a₂ – a₁ = 8

Therefore, the series is in AP

Here, a = 120, d = 8 and a_n = 496

We know that,

a_n = a + (n – 1)d

⇒ 496 = 104 + (n – 1)8

⇒ 496 – 104 = (n – 1)8

⇒ 392 = (n – 1)8

⇒ 49 = (n – 1)

⇒ n = 50

Now, we have to find the sum of this AP

⇒ S₅₀ = 25[208 + 49 × 8]

⇒ S₅₀ = 25[600]

⇒ S₅₀ = 15000

Hence, the sum of all numbers lying between 100 and 500 and divisible by 8 is 15000.