Arithmetic Progressions

Given,

a_n = n² – n+1

We can find first five terms of this sequence by putting values of n from 1 to 5.

When n = 1:

a₁ = (1)² – 1 + 1

⇒ a₁ = 1 – 1 + 1

⇒ a₁ = 1

When n = 2:

a₂ = (2)² – 2 + 1

⇒ a₂ = 4 – 2 + 1

⇒ a₂ = 3

When n = 3:

a₃ = (3)² – 3 + 1

⇒ a₃ = 9 – 3 + 1

⇒ a₃ = 7

When n = 4:

a₄ = (4)² – 4 + 1

⇒ a₄ = 16 – 4 + 1

⇒ a₄ = 13

When n = 5:

a₅ = (5)² – 5 + 1

⇒ a₅ = 25 – 5 + 1

⇒ a₅ = 21

∴ First five terms of the sequence are 1, 3, 7, 13, 21.

Given,

a_n = n³ – 6n² + 11n – 6, n ∈ N

We can find first three terms of sequence by putting the values of n form 1 to 3.

When n = 1:

a₁ = (1)³ – 6(1)² + 11(1) – 6

⇒ a₁ = 1 – 6 + 11 – 6

⇒ a₁ = 12 – 12

⇒ a₁ = 0

When n = 2:

a₂ = (2)³ – 6(2)² + 11(2) – 6

⇒ a₂ = 8 – 6(4) + 22 – 6

⇒ a₂ = 8 – 24 + 22 – 6

⇒ a₂ = 30 – 30

⇒ a₂ = 0

When n = 3:

a₃ = (3)³ – 6(3)² + 11(3) – 6

⇒ a₃ = 27 – 6(9) + 33 – 6

⇒ a₃ = 27 – 54 + 33 – 6

⇒ a₃ = 60 – 60

⇒ a₃ = 0

This shows that the first three terms of the sequence is zero.

When n = n:

a_n = n³ – 6n² + 11n – 6

⇒ a_n = n³ – 6n² + 11n – 6 – n + n – 2 + 2

⇒ a_n = n³ – 6n² + 12n – 8 – n + 2

⇒ a_n = (n)³ – 3×2n(n – 2) – (2)³ – n + 2

{(a – b)³ = (a)³ – (b)³ – 3ab(a – b)}

⇒ a_n = (n – 2)³ – (n – 2)

Here, n – 2 will always be positive for n > 3

∴ a_n is always positive for n > 3

Given,

a₁ = 3 and a_n = 3a_n–1 + 2, for all n > 1

We can find the first four terms of a sequence by putting values of n

from 1 to 4

When n = 1:

a₁ = 3

When n = 2:

a₂ = 3a_2–1 + 2

⇒ a₂ = 3a₁ + 2

⇒ a₂ = 3(3) + 2

⇒ a₂ = 9 + 2

⇒ a₂ = 11

When n = 3:

a₃ = 3a_3–1 + 2

⇒ a₃ = 3a₂ + 2

⇒ a₃ = 3(11) + 2

⇒ a₃ = 33 + 2

⇒ a₃ = 35

When n = 4:

a₄ = 3a_4–1 + 2

⇒ a₄ = 3a₃ + 2

⇒ a₄ = 3(35) + 2

⇒ a₄ = 105 + 2

⇒ a₄ = 107

∴ First four terms of sequence are 3, 11, 35, 107.

Given,

a₁ = 1, a_n = a_n–1 + 2, n > 1

We can find the first five terms of a sequence by putting values of n

from 1 to 5

When n = 1:

a₁ = 1

When n = 2:

a₂ = a_2–1 + 2

⇒ a₂ = a₁ + 2

⇒ a₂ = 1 + 2

⇒ a₂ = 3

When n = 3:

a₃ = a_3–1 + 2

⇒ a₃ = a₂ + 2

⇒ a₃ = 3 + 2

⇒ a₃ = 5

When n = 4:

a₄ = a_4–1 + 2

⇒ a₄ = a₃ + 2

⇒ a₄ = 5 + 2

⇒ a₄ = 7

When n = 5:

a₅ = a_5–1 + 2

⇒ a₅ = a₄ + 2

⇒ a₅ = 7 + 2

⇒ a₅ = 9

∴ First five terms of the sequence are 1, 3, 5, 7, 9.

Given,

a₁ = 1 = a₂, a_n = a_n–1 + a_n–2, n>2

We can find the first five terms of a sequence by putting values of n

from 1 to 5

When n = 1:

a₁ = 1

When n = 2:

a₁ = 1

When n = 3:

a₃ = a_3–1 + a_3–2

⇒ a₃ = a₂ + a₁

⇒ a₃ = 1 + 1

⇒ a₃ = 2

When n = 4:

a₄ = a_4–1 + a_4–2

⇒ a₄ = a₃ + a₂

⇒ a₄ = 2 + 1

⇒ a₄ = 3

When n = 5:

a₅ = a_5–1 + a_5–2

⇒ a₅ = a₄ + a₃

⇒ a₅ = 3 + 2

⇒ a₅ = 5

∴ First five terms of the sequence are 1, 1, 2, 3, 5.

Given,

a₁ = 2 = a₂, a_n = a_n–1 – 1, n>2

We can find the first five terms of a sequence by putting values of n

from 1 to 5

When n = 1:

a₁ = 2

When n = 2:

a₁ = 2

When n = 3:

a₃ = a_3–1 – 1

⇒ a₃ = a₂ – 1

⇒ a₃ = 2 – 1

⇒ a₃ = 1

When n = 4:

a₄ = a_4–1 – 1

⇒ a₄ = a₃ – 1

⇒ a₄ = 1 – 1

⇒ a₄ = 0

When n = 5:

a₅ = a_5–1 – 1

⇒ a₅ = a₄ – 1

⇒ a₅ = 0 – 1

⇒ a₅ = -1

∴ First five terms of the sequence are 2, 2, 1, 0, -1.

Given: a₁ = 1 = a₂, a_n = a_n–1 + a_n–2, n>2

When n = 1:

a₃ = a_3–1 + a_3–2

⇒ a₃ = a₂ + a₁

⇒ a₃ = 1 + 1

⇒ a₃ = 2

When n = 2:

a₄ = a_4–1 + a_4–2

⇒ a₄ = a₃ + a₂

⇒ a₄ = 2 + 1

⇒ a₄ = 3

When n = 3:

a₅ = a_5–1 + a_5–2

⇒ a₅ = a₄ + a₃

⇒ a₅ = 3 + 2

⇒ a₅ = 5

When n = 4:

a₆ = a_6–1 + a_6–2

⇒ a₆ = a₅ + a₄

⇒ a₆ = 5 + 3

⇒ a₆ = 8

When n = 5:

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a₁ = 3, a₂ = -1, a₃ = -5, a₄ = -9

Now, a₂ – a₁ = -1 – 3 = -4

a₃ – a₂ = -5 – (-1) = -5 + 1 = -4

a₄ – a₃ = -9 – (-5) = -9 + 5 = -4

As, a₂ – a₁ = a₃ – a₂ = a₄ – a₃

The given sequence is A.P

Common difference, d = a₂ – a₁ = -4

To find next three more terms of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is first term or a₁ and d is common difference

∴ a_n = 3 + (n-1) -4

⇒ a_n = 3 – 4n + 4

⇒ a_n = 7 – 4n

When n = 5:

a₅ = 7 – 4(5)

⇒ a₅ = 7 – 20

⇒ a₅ = -13

When n = 6:

a₆ = 7 – 4(6)

⇒ a₆ = 7 – 24

⇒ a₆ = -17

When n = 7:

a₇ = 7 – 4(7)

⇒ a₇ = 7 – 28

⇒ a₇ = -21

Hence, A.P is 3, -1, -5, -9, -13, -17, -21,…

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

As, a₂ – a₁ = a₃ – a₂ = a₄ – a₃

The given sequence is A.P

Common difference, d

To find next three more terms of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is first term or a₁ and d is common difference

When n = 5:

⇒ a₅ = 4

When n = 6:

When n = 7:

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

As, a₂ – a₁ = a₃ – a₂ = a₄ – a₃

The given sequence is A.P

Common difference,d

To find the next three more terms of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is a first term or a₁ and d is common difference

When n = 5:

When n = 6:

When n = 7:

Hence, A.P. is

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a₁ = 9, a₂ = 7, a₃ = 5, a₄ = 3

Now, a₂ – a₁ = 7 – 9 = -2

a₃ – a₂ = 5 – 7 = -2

a₄ – a₃ = 3 – 5 = -2

As, a₂ – a₁ = a₃ – a₂ = a₄ – a₃

The given sequence is A.P

Common difference, d = a₂ – a₁ = -2

To find the next three more terms of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is first term or a₁ and d is common difference

∴ a_n = 9 + (n-1) -2

⇒ a_n = 9 – 2n + 2

⇒ a_n = 11 – 2n

When n = 5:

a₅ = 11 – 2(5)

⇒ a₅ = 11 – 10

⇒ a₅ = 1

When n = 6:

a₆ = 11 – 2(6)

⇒ a₆ = 11 – 12

⇒ a₆ = -1

When n = 7:

a₇ = 11 – 2(7)

⇒ a₇ = 11 – 14

⇒ a₇ = -3

Hence, A.P is 9, 7, 5, 3, 1, -1, -3,….

Given,

a_n = 2n + 7

We can find first five terms of this sequence by putting values of n from 1 to 5.

When n = 1:

a₁ = 2(1) + 7

⇒ a₁ = 2 + 7

⇒ a₁ = 9

When n = 2:

a₂ = 2(2) + 7

⇒ a₂ = 4 + 7

⇒ a₂ = 11

When n = 3:

a₃ = 2(3) + 7

⇒ a₃ = 6 + 7

⇒ a₃ = 13

When n = 4:

a₄ = 2(4) + 7

⇒ a₄ = 8 + 7

⇒ a₄ = 15

When n = 5:

a₅ = 2(5) + 7

⇒ a₅ = 10 + 7

⇒ a₅ = 17

∴ First five terms of the sequence are 9, 11, 13, 15, 17.

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a₁ = 9, a₂ = 11, a₃ = 13, a₄ = 15, a₅ = 17

Now, a₂ – a₁ = 11 – 9 = 2

a₃ – a₂ = 13 – 11 = 2

a₄ – a₃ = 15 – 13 = 2

a₅ – a₄ = 17 – 15 = 2

As, a₂ – a₁ = a₃ – a₂ = a₄ – a₃ = a₅ – a₄

The given sequence is A.P

Common difference, d = a₂ – a₁ = 2

To find the seventh term of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is first term or a₁ and d is common difference

∴ a_n = 3 + (n-1) 2

⇒ a_n = 3 + 2n – 2

⇒ a_n = 2n + 1

When n = 7:

a₇ = 2(7) + 1

⇒ a₇ = 14 + 1

⇒ a₇ = 15

Hence, the 7^th term of A.P. is 15

Given,

a_n = 2n² + n + 1

We can find first three terms of this sequence by putting values of n from 1 to 3.

When n = 1:

a₁ = 2(1)² + 1 + 1

⇒ a₁ = 2(1) + 2

⇒ a₁ = 2 + 2

⇒ a₁ = 4

When n = 2:

a₂ = 2(2)² + 2 + 1

⇒ a₂ = 2(4) + 3

⇒ a₂ = 8 + 3

⇒ a₂ = 11

When n = 3:

a₃ = 2(3)² + 3 + 1

⇒ a₃ = 2(9) + 4

⇒ a₃ = 18 + 4

⇒ a₃ = 22

∴ First three terms of the sequence are 4, 11, 22.

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a₁ = 4, a₂ = 11, a₃ = 22

Now, a₂ – a₁ = 11 – 4 = 7

a₃ – a₂ = 22 – 11 = 11

As a₂ – a₁ is not equal to a₃ – a₂

The given sequence is not an A.P.

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a = a₁ = 1, a₂ = 4

Common difference, d = a₂ – a₁ = 4 – 1 = 3

To find the tenth term of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is first term or a₁ and d is common difference

∴ a_n = 1 + (n-1) 3

⇒ a_n = 1 + 3n – 3

⇒ a_n = 3n – 2

When n = 10:

a₁₀ = 3(10) – 2

⇒ a₁₀ = 30 – 2

⇒ a₁₀ = 28

Hence, 10^th term of A.P. is 28

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

Common difference, d

To find 18^th term of A.P, firstly find a_n

We know, a_n = a + (n-1) d where a is a first term or a₁ and d is common difference

When n = 18:

Hence, 18th term of A.P. is 34√2

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a = a₁ = 13, a₂ = 8

Common difference, d = a₂ – a₁ = 8 – 13 = -5

We know, a_n = a + (n-1) d where a is first term or a₁ and d is common difference

∴ a_n = 13 + (n-1) -5

⇒ a_n = 13 – 5n + 5

⇒ a_n = 18 – 5n

Hence, n^th term of A.P. is 18 – 5n

Let common difference of an A.P is d and first term is a

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference

Now, take L.H.S.:

a_m+n + a_m-n = a + (m + n – 1)d + a + (m - n – 1)d

⇒ a_m+n + a_m-n = a + md + nd – d + a + md - nd – d

⇒ a_m+n + a_m-n = 2a + 2md – 2d

⇒ a_m+n + a_m-n = 2(a + md – d)

⇒ a_m+n + a_m-n = 2[a + d(m – 1)]

{∵ a_n = a + (n – 1)d}

⇒ a_m+n + a_m-n = 2a_m

Hence Proved

Given A.P is 3, 8, 13,…

Here, a₁ = a = 3, a₂ = 8

Common difference, d = a₂ – a₁ = 8 – 3 = 5

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference

∴ a_n = 3 + (n – 1)5

⇒ a_n = 3 + 5n – 5

⇒ a_n = 5n – 2

Now, to find which term of A.P is 248

Put a_n = 248

∴ 5n – 2 = 248

⇒ 5n = 248 + 2

⇒ 5n = 250

⇒ n = 50

Hence, 50^th term of given A.P is 248

Given A.P is 84, 80, 76,…

Here, a₁ = a = 84, a₂ = 88

Common difference, d = a₂ – a₁ = 80 – 84 = -4

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference

∴ a_n = 84 + (n – 1)-4

⇒ a_n = 84 – 4n + 4

⇒ a_n = 88 – 4n

Now, to find which term of A.P is 0

Put a_n = 0

∴ 88 – 4n = 0

⇒ -4n = -88

⇒ n = 22

Hence, 22^th term of given A.P is 0

Given A.P is 4, 9, 14,…

Here, a₁ = a = 4, a₂ = 9

Common difference, d = a₂ – a₁ = 9 – 4 = 5

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference

∴ a_n = 4 + (n – 1)5

⇒ a_n = 4 + 5n – 5

⇒ a_n = 5n – 1

Now, to find which term of A.P is 254

Put a_n = 254

∴ 5n – 1 = 254

⇒ 5n = 254 + 1

⇒ 5n = 255

⇒ n = 51

Hence, 51^st term of given A.P is 254

Given A.P is 7, 10, 13,…

Here, a₁ = a = 7, a₂ = 10

Common difference, d = a₂ – a₁ = 10 – 7 = 3

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

∴ a_n = 7 + (n – 1)3

⇒ a_n = 7 + 3n – 3

⇒ a_n = 3n + 4

Now, to find whether 68 is a term of this A.P. or not

Put a_n = 68

∴ 3n + 4 = 68

⇒ 3n = 68 – 4

⇒ 3n = 64

is not a natural number

Hence, 68 is not a term of given A.P.

Given A.P is 3, 8, 13,…

Here, a₁ = a = 3, a₂ = 8

Common difference, d = a₂ – a₁ = 8 – 3 = 5

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

∴ a_n = 3 + (n – 1)5

⇒ a_n = 3 + 5n – 5

⇒ a_n = 5n – 2

To find: whether 302 is a term of this A.P. or not

Put a_n = 302

∴ 5n – 2 = 302

⇒ 5n = 302 + 2

⇒ 5n = 304

is not a natural number

Hence,304 is not a term of given A.P.

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

Now, to find first negative term

Put a_n < 0

⇒ n > 33

Hence, 34^th term is first negative term of given A.P

Given A.P is 12 + 8i, 11 + 6i, 10 + 4i, …

Here, a₁ = a = 12 + 8i, a₂ = 11 + 6i

Common difference, d = a₂ – a₁

= 11 + 6i – (12 + 8i) = 11 – 12 + 6i – 8i = -1 – 2i

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

∴ a_n = 12 + 8i + (n – 1) -1 – 2i

⇒ a_n = 12 + 8i – n – 2ni + 1 + 2i

⇒ a_n = 13 + 10i – n – 2ni

⇒ a_n = (13 – n) + (10 – 2n)i

To find purely real term of this A.P., imaginary part have to be zero

∴ 10 – 2n = 0

⇒ 2n = 10

⇒ n = 5

Hence, 5^th term is purely real

To find purely imaginary term of this A.P., real part have to be zero

∴ 13 – n = 0

⇒ n = 13

Hence, 13^th term is purely imaginary

Given A.P is 7, 10, 13,…

Here, a₁ = a = 7, a₂ = 10

Common difference, d = a₂ – a₁ = 10 – 7 = 3

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

∴ a_n = 7 + (n – 1)3

⇒ a_n = 7 + 3n – 3

⇒ a_n = 3n + 4

To find total terms of the A.P., put a_n = 43 as 43 is last term of A.P.

∴ 3n + 4 = 43

⇒ 3n = 43 – 4

⇒ 3n = 39

⇒ n = 13

Hence, there are total 13 terms in the given A.P.

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

To find total terms of the A.P., put a_{_n}is last term of A.P.

⇒ n = 27

Hence, there are total 27 terms in the given A.P.

Given,

a = 5, last term, l = a_n = 80

Common difference, d = 3

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

∴ a_n = 5 + (n – 1)3

⇒ a_n = 5 + 3n – 3

⇒ a_n = 3n + 2

To find total terms of the A.P., put a_n = 80 as 80 is last term of A.P.

∴ 3n + 2 = 80

⇒ 3n = 80 – 2

⇒ 3n = 78

⇒ n = 26

Hence, there are total 26 terms in the given A.P.

Given,

6^th term of an A.P is 19 and 17^th terms of an A.P. is 41

⇒ a₆ = 19 and a₁₇ = 41

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

Similarly, When n = 17:

∴ a₁₇ = a + (17 – 1)d

⇒ a₁₇ = a + 16d

According to question:

a₆ = 19 and a₁₇ = 41

⇒ a + 5d = 19 ………………(i)

And a + 16d = 41…………..(ii)

Subtracting equation (i) from (ii):

a + 16d – (a + 5d) = 41 – 19

⇒ a + 16d – a – 5d = 22

⇒ 11d = 22

⇒ d = 2

Put the value of d in equation (i):

a + 5(2) = 19

⇒ a + 10 = 19

⇒ a = 19 – 10

⇒ a = 9

As, a_n = a + (n – 1)d

a₄₀ = a + (40 – 1)d

⇒ a₄₀ = a + 39d

Now put the value of a = 9 and d = 2 in a₄₀

⇒ a₄₀ = 9 + 39(2)

⇒ a₄₀ = 9 + 78

⇒ a₄₀ = 87

Hence, 40^th term of the given A.P. is 87

Given,

9^th term of an A.P is 0

⇒ a₉ = 0

To prove: a₂₉ = 2a₁₉

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 9:

∴ a₉ = a + (9 – 1)d

⇒ a₉ = a + 8d

According to question:

a₉ = 0

⇒ a + 8d = 0

⇒ a = -8d

When n = 19:

∴ a₁₉ = a + (19 – 1)d

⇒ a₁₉ = a + 18d

⇒ a₁₉ = -8d + 18d

⇒ a₁₉ = 10d

When n = 29:

∴ a₂₉ = a + (29 – 1)d

⇒ a₂₉ = a + 28d

{∵ a = -8d}

⇒ a₂₉ = -8d + 28d

⇒ a₂₉ = 20d

⇒ a₂₉ = 2×10d

{∵ a₁₉ = 10d}

⇒ a₂₉ = 2a₁₉

Hence Proved

Given,

10 times the 10^th term of an A.P. is equal to 15 times the 15^th term

⇒ 10a₁₀ = 15a₁₅

To prove: a₂₅ = 0

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

When n = 15:

∴ a₁₅ = a + (15 – 1)d

⇒ a₁₅ = a + 14d

When n = 25:

∴ a₂₅ = a + (25 – 1)d

⇒ a₂₅ = a + 24d ………(i)

According to question:

10a₁₀ = 15a₁₅

⇒ 10(a + 9d) = 15(a + 14d)

⇒ 10a + 90d = 15a + 210d

⇒ 10a – 15a + 90d – 210d = 0

⇒ -5a – 120d = 0

⇒ -5(a + 24d) = 0

⇒ a + 24d = 0

⇒ a₂₅ = 0 (From (i))

Hence Proved

Given: 10^th term of an A.P is 41, and 18^th terms of an A.P. is 73

⇒ a₁₀ = 41 and a₁₈ = 73

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is the common difference and n is any natural number

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

Similarly, When n = 18:

∴ a₁₈ = a + (18 – 1)d

⇒ a₁₈ = a + 17d

According to question:

a₁₀ = 41 and a₁₈ = 73

⇒ a + 9d = 41 ………………(i)

And a + 17d = 73…………..(ii)

Subtracting equation (i) from (ii):

a + 17d – (a + 9d) = 73 – 41

⇒ a + 17d – a – 9d = 32

⇒ 8d = 32

⇒ d = 4

Put the value of d in equation (i):

a + 9(4) = 41

⇒ a + 36 = 41

⇒ a = 41 – 36

⇒ a = 5

As, a_n = a + (n – 1)d

a₂₆ = a + (26 – 1)d

⇒ a₂₆ = a + 25d

Now put the value of a = 5 and d = 4 in a₂₆

⇒ a₂₆ = 5 + 25(4)

⇒ a₂₆ = 5 + 100

⇒ a₂₆ = 105

Hence, 26^th term of the given A.P. is 105

Given: 24^th term is twice the 10^th term

⇒ a₂₄ = 2a₁₀

To prove: a₇₂ = 2a₃₄

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

When n = 24:

∴ a₂₄ = a + (24 – 1)d

⇒ a₂₄ = a + 23d

When n = 34:

∴ a₃₄ = a + (34 – 1)d

⇒ a₃₄ = a + 33d ………(i)

When n = 72:

∴ a₇₂ = a + (72 – 1)d

⇒ a₇₂ = a + 71d

According to question:

a₂₄ = 2a₁₀

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ a – 2a + 23d – 18d = 0

⇒ -a + 5d = 0

⇒ a = 5d

Now, a₇₂ = a + 71d

⇒ a₇₂ = 5d + 71d

⇒ a₇₂ = 76d

⇒ a₇₂ = 10d + 66d

⇒ a₇₂ = 2(5d + 33d)

{∵ a = 5d}

⇒ a₇₂ = 2(a + 33d)

⇒ a₇₂ = 2a₃₄ (From (i))

Hence Proved

Given: (m + 1)^th term of an A.P. is twice the (n + 1)^th term

⇒ a_m+1 = 2a_n+1

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

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = m + 1:

∴ a_m+1 = a + (m + 1 – 1)d

⇒ a_m+1 = a + md

When n = n + 1:

∴ a_n+1 = a + (n + 1 – 1)d

⇒ a_n+1 = a + nd

According to question:

a_m+1 = 2a_n+1

⇒ a + md = 2(a + nd)

⇒ a + md = 2a + 2nd

⇒ a – 2a + md – 2nd = 0

⇒ -a + (m – 2n)d = 0

⇒ a = (m – 2n)d………(i)

a_n = a + (n – 1)d

When n = m + n + 1:

∴ a_m+n+1 = a + (m + n + 1 – 1)d

⇒ a_m+n+1 = a + md + nd

⇒ a_m+n+1 = (m – 2n)d + md + nd……… (From (i))

⇒ a_m+n+1 = md – 2nd + md + nd

⇒ a_m+n+1 = 2md – nd………(ii)

When n = 3m + 1:

∴ a_3m+1 = a + (3m + 1 – 1)d

⇒ a_3m+1 = a + 3md

⇒ a_3m+1 = (m – 2n)d + 3md……… (From (i))

⇒ a_3m+1 = md – 2nd + 3md

⇒ a_3m+1 = 4md – 2nd

⇒ a_3m+1 = 2(2md – nd)

⇒ a_3m+1 = 2a_m+n+1…………(From (ii))

Hence Proved

Given: n^th term of the A.P. 9, 7, 5,… is same as the nth term of the A.P. 15, 12, 9 …

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

Let A.P. 9, 7, 5,… has first term a₁ and common difference d₁

⇒ a₁ = 9 and a₂ = 7

Common difference, d₁ = a₂ – a₁ = 7 – 9 = -2

Now, a_n = a₁ + (n – 1)d₁

⇒ a_n = 9 + (n – 1)(-2)

⇒ a_n = 9 – 2n + 2

⇒ a_n = 11 – 2n

Let A.P. 15, 12, 9 … has first term a₁ and common difference d₁

⇒ b₁ = 15 and b₂ = 12

Common difference, d₂ = b₂ – b₁ = 12 – 15 = -3

Now, b_n = b₁ + (n – 1)d₂

⇒ b_n = 15 + (n – 1)(-3)

⇒ b_n = 15 – 3n + 3

⇒ b_n = 12 – 3n

According to question:

a_n = b_n

⇒ 11 – 2n = 12 – 3n

⇒ 3n – 2n = 12 – 11

⇒ n = 1

Hence, the value of n is 1

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a = a₁ = 3, a₂ = 5, l = 201

Common difference, d = a₂ – a₁ = 5 – 3 = 2

We know, n^th term from end, b_n = l – (n – 1)d where l is last term or a₁ and d is common difference and n is any natural number

∴ b₁₂ = 201 – (12 – 1)2

⇒ b₁₂ = 201 – 24 + 2

⇒ b₁₂ = 203 – 24

⇒ b₁₂ = 179

Hence, 12^thterm from end for the given A.P is 179

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a = a₁ = 3, a₂ = 8, l = 253

Common difference, d = a₂ – a₁ = 8 – 3 = 5

We know, n^th term from end, b_n = l – (n – 1)d where l is last term or a₁ and d is common difference and n is any natural number

∴ b₁₂ = 253 – (12 – 1)5

⇒ b₁₂ = 253 – 60 + 5

⇒ b₁₂ = 258 – 60

⇒ b₁₂ = 198

Hence, 12^thterm from end for the given A.P is 198

A.P is known for Arithmetic Progression whose common difference = a_n – a_n-1 where n > 0

a = a₁ = 1, a₂ = 4, l = 88

Common difference, d = a₂ – a₁ = 4 – 1 = 3

We know, n^th term from end, b_n = l – (n – 1)d where l is last term or a₁ and d is common difference and n is any natural number

∴ b₁₂ = 88 – (12 – 1)3

⇒ b₁₂ = 88 – 36 + 3

⇒ b₁₂ = 91 – 36

⇒ b₁₂ = 55

Hence, 12^thterm from end for the given A.P is 55

Given,

4^th term of an A.P. is three times the first and the 7^th term exceeds twice the third term by 1

⇒ a₄ = 3a and a₇ = 2a₃ + 1

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 4:

∴ a₄ = a + (4 – 1)d

⇒ a₄ = a + 3d

When n = 7:

∴ a₇ = a + (7 – 1)d

⇒ a₇ = a + 6d

When n = 3:

∴ a₃ = a + (3 – 1)d

⇒ a₃ = a + 2d

According to question:

a₇ = 2a₃ + 1

⇒ a + 6d = 2(a + 2d) + 1

⇒ a + 6d = 2a + 4d + 1

⇒ a – 2a + 6d – 4d = 1

⇒ -a + 2d = 1

⇒ a – 2d = -1……(i)

a₄ = 3a

⇒ a + 3d = 3a

⇒ 3d = 3a – a

⇒ 3d = 2a

Put this value of d in equation (i):

⇒ a = 3

Now, put this value of a in equation (ii):

⇒ d = 2

Hence, the value of a and d are 3 and 2 respectively

Given: 6^thterm of an A.P is 12 and 8^th terms of an A.P. is 22

⇒ a₆ = 12 and a₈ = 22

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

Similarly, When n = 8:

∴ a₈ = a + (8 – 1)d

⇒ a₈ = a + 7d

According to question:

a₆ = 12 and a₈ = 22

⇒ a + 5d = 12 ………………(i)

And a + 7d = 22…………..(ii)

Subtracting equation (i) from (ii):

a + 7d – (a + 5d) = 22 – 12

⇒ a + 7d – a – 5d = 10

⇒ 2d = 10

⇒ d = 5

Put the value of d in equation (i):

a + 5(5) = 12

⇒ a + 25 = 12

⇒ a = 12 – 25

⇒ a = -13

As, a_n = a + (n – 1)d

a₂ = a + (2 – 1)d

⇒ a₂ = a + d

Now put the value of a = 9 and d = 2 in a_n and a₂

⇒ a_n = a + (n – 1)d

⇒ a_n = -13 + (n – 1)5

⇒ a_n = -13 + 5n – 5

⇒ a_n = -18 + 5n

a₂ = a + d

⇒ a₂ = -13 + 5

⇒ a₂ = -8

Hence, 2^thterm and n^th of the given A.P. are -8 and 5n – 18 respectively

For finding total two-digit numbers which are divisible by 3, firstly we will make an A.P. of those two-digit numbers which are divisible by 3.

First two digit number which is divisible by 3 is 12

∴ a₁ = a = 12

Next two digit number which is divisible by 3 is 15

∴ a₂ = 15

Largest two digit number which is divisible by 3 is 99

∴ a_n = 99

⇒ A.P. is 12, 15,………,99

We know, a_n = a + (n – 1)d where a is first term or a and d is common difference and n is any natural number

⇒ a₁ = 12, a₂ = 15 and a_n = 99

Common difference, d₁ = a₂ – a₁ = 15 – 12 = 3

Now, a_n = a₁ + (n – 1)d

⇒ a_n = 12 + (n – 1)3

⇒ 99 = 12 + 3n – 3

⇒ 99 = 3n + 9

⇒ 99 – 9 = 3n

⇒ 90 = 3n

⇒ n = 30

Hence, there are total 30 two-digit numbers which are divisible by 3

Given: a = 7, n = 60 and l = a₆₀ = 125

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is common difference and n is any natural number

When n = 60:

∴ a₆₀ = a + (60 – 1)d

⇒ a₆₀ = a + 59d

When n = 32:

∴ a₃₂ = a + (32 – 1)d

⇒ a₃₂ = a + 31d …………(i)

According to question:

a₆₀ = 125

⇒ a + 59d = 125

⇒ 7 + 59d = 125

⇒ 59d = 125 – 7

⇒ 59d = 118

⇒ d = 2

Now put the value of a = 7 and d = 2 in a₃₂

⇒ a₃₂ = a + 31d………(From (i))

⇒ a₃₂ = 7 + 31(2)

⇒ a₃₂ = 7 + 62

⇒ a₃₂ = 69

Hence, 32^thterm of the given A.P. is 69

Given: the sum of the 4^th and the 8^th terms of an A.P. is 24, and the sum of the 6^th and 10^th term is 34

⇒ a₄ + a₈ = 24 and a₆ + a₁₀ = 34

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is the common difference and n is any natural number

When n = 4:

∴ a₄ = a + (4 – 1)d

⇒ a₄ = a + 3d

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

When n = 8:

∴ a₈ = a + (8 – 1)d

⇒ a₈ = a + 7d

When n = 10:

∴ a₁₀ = a + (10 – 1)d

⇒ a₁₀ = a + 9d

According to question:

a₄ + a₈ = 24

⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d = 24

⇒ 2(a + 5d) = 24

⇒ a + 5d = 12………(i)

a₆ + a₁₀ = 34

⇒ a + 5d + a + 9d = 34

⇒ 2a + 14d = 34

⇒ 2(a + 7d) = 34

⇒ a + 7d = 17………(ii)

Subtracting equation (i) from (ii):

a + 7d – (a + 5d) = 17 – 12

⇒ a + 7d – a – 5d = 5

⇒ 2d = 5

Put the value of d in equation (i):

Hence, the value of a and d areand respectively

For finding total numbers between 1 and 1000 which when divided by 7 leave remainder 4, firstly we will make an A.P. of those numbers which when divided by 7 leave remainder 4.

First number which when divided by 7 leave remainder 4 is 4

∴ a₁ = a = 4

Next number which when divided by 7 leave remainder 4 is 11

∴ a₂ = 11

Largest three-digit number which when divided by 7 leave remainder 4 is 998

∴ a_n = 998

⇒ A.P. is 4, 11,………,998

We know, a_n = a + (n – 1)d where a is first term or a and d is common difference and n is any natural number

⇒ a₁ = 4, a₂ = 11 and a_n = 998

Common difference, d₁ = a₂ – a₁ = 11 – 4 = 7

Now, a_n = a₁ + (n – 1)d

⇒ a_n = 4 + (n – 1)7

⇒ 998 = 4 + 7n – 7

⇒ 998 = 7n – 3

⇒ 998 + 3 = 7n

⇒ 1001 = 7n

⇒ n = 143

Hence, there are total 143 numbers between 1 and 1000 which when divided by 7 leave remainder 4.

To prove: sum of the n^th term from the beginning and the nth term from the end is a + l

We know, a_n = a + (n – 1)d where a is first term or a₁ and d is the common difference, and n is any natural number, and n^th term from the end is a_n^’ = l – (n – 1)d

Now,

a_n + a_n^’ = a + (n – 1)d + l – (n – 1)d

⇒ a_n + a_n^’ = a + nd – d + l – nd + d

⇒ a_n + a_n^’ = a + l

Hence Proved

Given:

To find:

We know, a_n = a + (n – 1)d where a is a first term or a₁ and d is the common difference and n is any natural number

When n = 4:

∴ a₄ = a + (4 – 1)d

⇒ a₄ = a + 3d

When n = 6:

∴ a₆ = a + (6 – 1)d

⇒ a₆ = a + 5d

When n = 7:

∴ a₇ = a + (7 – 1)d

⇒ a₇ = a + 6d

When n = 8:

∴ a₈ = a + (8 – 1)d

⇒ a₈ = a + 7d

According to the question:

⇒ 3(a + 3d) = 2(a + 6d)

⇒ 3a + 9d = 2a + 12d

⇒ 3a – 2a = 12d – 9d

⇒ a = 3d

Now,

Hence, the value of

Given: θ₁, θ₂, θ₃,…, θ_n is A.P

∴ θ₂ – θ₁ = θ₃ – θ₂ = θ₄ – θ₃ =…………= θ_n – θ_n-1 = d (Common Difference)

Now,

Multiplying and dividing by sin d:

{∵ sin(A - B) = sin A cos B - cos A sin B}

Similarly,

Take L.H.S.:

Putting the value of (i), (ii) and (iii):

= R.H.S.

Hence Proved

Given: the sum of first three terms is 21

To find: the first three terms of AP

Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference

So, sum of first three terms is a - d + a + a + d = 21

3a = 21

a = 7

it is also given that product of first and third term exceeds the second by 6

so (a - d)(a + d) - a = 6

a² - d² - a = 6

substituting a = 7

7² - d² - 7 = 6

d² = 36

d = 6 or d = - 6

Hence the terms of AP are a - d, a, a + d which is 1, 7, 13 or 13, 7, 1

Given: sum of first three terms is 27

To find: the first three terms of AP

Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference

So, sum of first three terms is a - d + a + a + d = 27

3a = 27

a = 9

given that the product of three terms is 648

so a³ - ad² = 648

substituting a = 9

9³ - 9d² = 648

729 - 9d² = 648

81 = 9d²

d = 3 or d = - 3

hence the given terms are a - d, a, a + d which is 6, 9, 12 or 12, 9, 6

Given: Sum of four terms is 50

To find: first four terms of AP

Assume these four terms are a - 2d, a - d, a + d, a + 2d

Sum of these terms is 4a = 50

It is also given that the greatest number is 4 time the least

a + 2d = 4(a - 2d)

3a = 10d

Substituting

Hence the terms of AP are a - d, a, a + d which is

assume the numbers in AP are a - d, a, a + d

Given that the sum of three numbers is 12

To find: the first three terms of AP

So,

3a = 12

a = 4

It is also given that the sum of their cube is 288

(a - d)³ + a³ + (a + d)³ = 288

a³ - d³ - 3ad(a - d) + a³ + a³ + d³ + 3ad(a + d) = 288

substituting a = 4 we get

64 - d³ - 12d(4 - d) + 64 + 64 + d³ + 12d(4 + d) = 288

192 + 24d² = 288

d = 2 or d = - 2

hence the numbers are a - d, a, a + d which is 2, 4, 6 or 6, 4, 2

Given: sum of first three terms is 24

To find: the first three terms of AP

Assume the first three terms are a - d, a, a + d where a is the first term and d is the common difference

So, sum of first three terms is a - d + a + a + d = 24

3a = 24

a = 8

given that the product of three terms is 440

so a³ - ad² = 440

substituting a = 8

8³ - 8d² = 440

512 - 8d² = 440

72 = 8d²

d = 3 or d = - 3

hence the given terms are a - d, a, a + d which is 5, 8, 11 or 11, 8, 5

assume the angles are a - 2d, a - d, a + d, a + 2d

We know that the sum of all angles in a quadrilateral is 360

Given: d = 10

To find: angles of the quadrilateral

So, a - 2d + a - d + a + d + a + 2d = 360

4a = 360

a = 90

hence the angles are a - 2d, a - d, a + d, a + 2d which is 70, 80, 100, 110

for the given AP the first term a is 50, and common difference d is a difference of second term and first term, which is 46 - 50 = - 4

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

s = 5 × 64

s = 320

for the given AP the first term a is 1, and common difference d is a difference of the second term and first term, which is 3 - 1 = 2

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

s = 6 × 24

s = 144

for the given AP the first term a is 3, and common difference d is a difference of the second term and first term, which is

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

s = 25 × 21

s = 525

for the given AP the first term a is 41, and common difference d is a difference of the second term and first term, which is 36 - 41 = - 5

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

s = 6 × 27

s = 162

for the given AP the first term a is a + b, and common difference d is a difference of the second term and the first term, which is a - b - (a + b) = - 2b

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

s = 11(2a - 40b)

for the given AP the first term a is (x - y)² and common difference d is a difference of the second term and first term, which is

To find: the sum of given AP

Formula: for the sum of AP is given by

Substituting the values in the above formula

for the given AP the first term a is (x - y)² and common difference d is a difference of the second term and first term, which is

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

for the given AP the first term a is 2, and common difference d is a difference of the second term and first term, which is 5 - 2 = 3

To find: the sum of given AP

The formula for the sum of AP is given by

Substituting the values in the above formula

s = 92 × 61

s = 5612

for the given AP the first term a is 101, and common difference d is a difference of second term and first term, which is 99 - 101 = - 2

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

s = 14 × 148

s = 2072

for the given AP the first term a is (a - b)² and common difference d is a difference of second term and first term, which is

To find: the sum of given AP

The formula for sum of AP is given by

Substituting the values in the above formula

n = 5

for the given AP the first term a is 1, and common difference d is a difference of the second term and first term, which is 2 - 1 = 1

To find: the sum of given AP

The formula for the sum of AP is given by

Substituting the values in the above formula

According to the given question we have to find the sum of natural numbers between 1 and 100 which are divisible by 2 or 5, so the sequence is 2, 4, 6…98 which are multiples of 2 and another sequence 5, 15, 25….95

To find: the sum of all - natural numbers between 1 and 100, which are divisible by 2 or 5

For the first sequence, it is an AP with first term a as 2 and common difference d as 2

Hence the sum is given by the formula

For the second sequence, it is an AP with first term a as 5 and common difference d as 10

Hence the sum is given by the formula

Hence total sum is s₁ + s₂ = 2450 + 500 = 2950

given an AP of first n odd natural numbers whose first term a is 1, and common difference d is 3

Given sequence is 1, 3, 5, 7……n

To find: the sum of first n natural numbers

Hence the sum is given by the formula

Substituting the values in the above equation we get

s = n²

the given sequence is 101, 103, ….199

Given an AP whose first term a is 101, and common difference d is 2

To find: the sum of all odd numbers between 100 and 200

Hence the sum is given by the formula

Substituting the values in the above equation we get

Now for the finding number of terms, the formula is

Substituting n is the sum formula we get

s = 25 × 300

s = 7500

according to given conditions the sequence is 3, 9, 15, 21….999

Given an AP whose first term is 3 and d is 6. Hence sum is given by

To prove: the sum of all odd integers between 1 and 1000 which are divisible by 3 is 83667

Hence the sum is given by the formula

Now for the finding number of terms, the formula is

n = 167

Substituting n is the sum formula we get

s = 83667

Hence, Proved.

given an AP is required of all integers between 84 and 719, which are multiples of 5

To find: the sum of all integers between 84 and 719, which are multiples of 5

So, the sequence is 85, 90, 95….715

It is an AP whose first term is 85 and d is 5

Hence the sum is given by the formula

Now for the finding number of terms, the formula is

n = 127

Substituting n is the sum formula we get

s = 50800

given an AP is required of all integers between 50 and 500, which are multiples of 7

To find: the sum of all integers between 50 and 500 which are divisible by 7

So, the sequence is 56, 63, 70….497

It is an AP whose first term is 56 and d is 7

Hence the sum is given by the formula

Now for the finding number of terms, the formula is

n = 64

Substituting n is the sum formula we get

s = 17696

given an AP is required of all integers between 101 and 999 which are even

To find: the sum of all even integers between 101 and 999

So, the sequence is 102, 104, 106….998

It is an AP whose first term is 102 and d is 2

Hence the sum is given by the formula

Now for the finding number of terms, the formula is

n = 449

Substituting n is the sum formula we get

s = 246950

given an AP is required of all integers between 100 and 550, which are multiples of 9

To find: the sum of all integers between 100 and 550, which are divisible by 9

So, the sequence is 108, 117, 126….549

It is an AP whose first term is 108 and d is 9

Hence the sum is given by the formula

Now for the finding number of terms, the formula is

n = 50

substituting n in the sum formula we get

s = 16425

given an AP is whose first term is 3 and d is 2

So, the sum is given by formula

To find: sum of the given AP

Substituting the values in the formula

s = 3n(3n + 2)

given an AP is required of all integers between 100 and 800, which on division by 16 leaves remainder 7

To find: the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7

So, the sequence is 103, 119, 135….791

It is an AP whose first term is 103 and d is 16

Hence the sum is given by the formula

Now for the finding number of terms, the formula is

n = 44

substituting n in the sum formula, we get

s = 19668

given is an AP whose first term a is 25 and d is - 3(22 - 25) with a number of terms equal to..

To find: the value of x for the given AP

Now for the finding number of terms, the formula is

Hence the sum is given by the formula

Now substituting the values in the above formula

It is given that s = 115, so equating the above expression to 115

Solving we get n is 10 or

Since n cannot be a fraction, so we choose n as 10

x = - 2

given is an AP whose first term is 1 and d is 3(4 - 1) with a number of terms equal to

To find: the value of x for the given AP

Now for the finding number of terms, the formula is

Hence the sum is given by the formula

Now substituting the values in the above formula

It is given that s = 590, so equating the above expression to 590

Solving we get n is 20 or

Since n cannot be a fraction, so we choose n as 20

x = 58

It is given that the sum of n terms is 3n² + 2n

To find: rth term of an AP, the sum of whose first term is 3n² + 2n

So, the sum of n - 1 terms is 3(n - 1)² + 2(n - 1)

formula T_N = S_N - S_{N - 1}

So, T_r = 3r² + 2r - [3(r - 1)² + 2(r - 1)]

T_r = 3r² + 2r - [3(r² + 1 - 2r) + 2(r - 1)]

T_r = - 1 + 6r

given the first term is - 14, from which we can say that a = - 14

To find: the number of terms in an AP whose first and fifth terms is - 14 and 2 respectively, and the sum of the terms is 40

Given fifth term is 2, so a + 4d = 2

- 14 + 4d = 2

d = 4

we know that the sum of AP is given by the formula:

now substituting the values in the above equation

4n² - 32n - 80 = 0

n² - 8n - 20 = 0

solving we get n as 10 or - 2

Since the number of terms cannot be negative hence n is 10

Assuming the first term as a and common difference as d

To find: the progression

So, the sum of first 7 terms is given by

a + a + d + a + 2d + a + 3d…. a + 6d = 10

7a + 21d = 10…(i)

In the second part it is given that sum of next seven terms is 17

a + 7d + a + 8d + a + 9d…. a + 13d = 7

7a + 70d = 7…(ii)

Solving (i) and (ii) we get

10 - 21d = 7 - 70d

3 = - 49d

Hence the sequence is given by a, a + d, a + 2d…. which is

given third term of AP is 7

So, a + 2d = 7…(i)

It is also given that the seventh term exceeds 3 times the third term by 2

To find: first term, the common difference and the sum of the first 20 terms

a + 6d - 3(a + 2d) = 2

- 2a = 2

a = - 1

d = 4

we know that the sum of AP is given by the formula:

substituting the values in above equation

s = 740

given a₁ = 2 and a_n = 50

To find: the common difference

we know that the sum of AP is given by the formula:

substituting the values in above equation we get

n = 17

Now for the finding number of terms, the formula is

we know that

d = 3

let the number of terms be 2n with first term a and common difference d

To find: number of terms and the series

The last term is given as a + (2n - 1) d

It is given that the last term exceeds the first by

So

…(i)

Sum of odd terms is 24

…(ii)

Sum of even terms is 30

…(iii)

Solving (i), (ii) and (iii) we get

d = 1.5

substituting d in (i) we get n = 4

and substituting d and n in (ii) we get a = 1.5

so, the number of terms is 2n = 8, and the sequence is given by a, a + d, a + 2d….

1.5, 3, 4.5, 6, 7.5 , 9, 10.5 , 12

given an AP whose sum of n terms is n²p and same AP with m terms whose sum is m²p

To prove: S_p = p³

we know that the sum of AP is given by the formula:

substituting the values in the above equation, we get

…. (i)

Similarly, for series with m terms

…(ii)

Subtracting (ii) from (i) we get

d = 2p

substituting d in (i) we get

a = p

Now using the sum formula for AP consisting of p terms we get

Substituting the values in the above equation

S_P = p³

assuming an AP whose first term is a and the common difference is d

To find: the sum of first 10 terms

Given 12^th term is - 13

So, a + 11d = - 13...(i)

Given the sum of first four terms is 24

So 4a + 6d = 24…(ii)

Solving (i) and (ii) we get

a = 9 and d = - 2

we know that the sum of AP is given by the formula:

substituting the values in the above equation

s = 0

assuming an AP whose first term is a and the common difference is d

To find: the sum of the first 20 terms

Given 5^th term is 30

So, a + 4d = 30...(i)

Given 12^th term is 65

So, a + 11d = 65…(ii)

Solving (i) and (ii) we get

a = 10 and d = 5

we know that the sum of AP is given by the formula:

substituting the values in the above equation

s = 1150

given an AP whose kth term is 5k + 1

To find: the sum of n terms of an AP whose kth term is 5k + 1

So substituting k = 1 to get the first term a₁ = 6

Substituting k = 2 to get the second term a₂ = 11

d = a₂ - a₁ = 5

we know that the sum of AP is given by the formula:

substituting the values in the above equation

the series which satisfies the above condition is

13, 17, 21….97

To find: the sum of all two - digit numbers which when divided by 4, yields 1 as the remainder

So, it is an AP whose first term is 13 and common difference d as 4

Now for the finding number of terms, the formula is

And

n = 22

we know that the sum of AP is given by the formula:

substituting the values in the above equation

s = 1210

given an AP whose first term is 25 and the common difference is - 3

To find: the last term of a given AP

we know that the sum of AP is given by the formula

Solving we get

3n² - 53n + 232 = 0

n = 8 or

n cannot be a fraction hence we choose n as 8

Now for the finding number of terms, the formula is

solving we get the last term as 4

given is an AP whose first term is 1 and d is 2

To find: the sum of odd integers from 1 to 2001

Now for the finding number of terms, the formula is

n = 1001

we know that the sum of AP is given by the formula:

substituting the values in the above equation

s = 1002001

given is an AP whose first term is - 6 and d is

To find: the number of terms required to make the sum of a given AP to - 25

we know that the sum of AP is given by the formula:

substituting the values in the above equation

100 = - 25n + n²

n² - 25n + 100 = 0

(n - 20) (n - 5) = 0

solving we get n = 20 or n = 5

given an AP whose first term is 2

To find: 20^th term of an AP

It is given that the sum of the first five terms is one-fourth of the next 5 terms

So

20a + 40d = 5a + 35d

15a + 5d = 0

Substituting a as 2

We get d = - 6

Hence 20^th term is a + 19d = - 112

To prove: S₁: S₂ = (2n + 1) : (n + 1)

we know that the sum of AP is given by the formula:

Substituting the values in the above equation

For the sum of odd terms, it is given by

s₂ = a₁ + a₃ + a₅ + …. a_{2n + 1}

s₂ = a + a + 2d + a + 4d + … + a + 2nd

s₂ = (n + 1)a + n(n + 1)d

s₂ = (n + 1) (a + nd)

Hence s₁:s₂

To find: AP with given conditions

Given that sum of n terms is 3n²

S_n = 3n²

Similarly, sum of n - 1 terms is 3(n - 1)²

S_{n - 1} = 3(n - 1)²

formula T_N = S_n - S_{n - 1} = 3n² - 3(n - 1)²

Now substituting n = 1 to get the first term

a₁ = 3

Now substituting n = 2 to get the second term

a₂ = 9

d = a₂ - a₁ = 6

hence the series is given by a, a + d, a + 2d…which is 3, 9, 15, 21….

it is given that sum of n terms of AP is

To find: the common difference

Substituting n = 1 gives the sum of the first term, that is the first term itself

a₁ = P

Substituting n = 2 gives the sum of first two terms

a₁ + a₂ = 2P + Q

a₂ = P + Q

Now common difference d = a₂ - a₁ = Q

assuming first AP whose first term is a and common difference d

To find: the ratio of 18^th term of a given AP

we know that the sum of AP is given by the formula:

Substituting the values in the above equation

assuming second AP whose first term is A and common difference D

we know that the sum of AP is given by the formula:

substituting the values in the above equation

…. (i)

Given the ratio of the 18^th term is required of both AP’s which is

Substituting n = 35 in (i) we get

assuming first AP whose first term is a and common difference d

To find: the ratio of 5^th term of a given AP

we know that the sum of AP is given by the formula:

substituting the values in the above equation

Assuming second AP whose first term is A and common difference D

we know that the sum of AP is given by the formula:

substituting the values in the above equation

…. (i)

Given the ratio of the 5^th term is required of both AP’s which is

Substituting n = 9 in (i) we get

We know, if a, b, c are in AP,

Then b - a = c - b

It is said that, are in AP,

Hence,

If are in AP, then

Taking LCM,

Now, LHS =

Multiply c in both numerator and denominator,

We get,

RHS

Multiply a in both numerator and denominator,

We get, =

Therefore, LHS = RHS

=

c(b - a) = a(b-c)

Also,

c(b - a) = a(b - c)

Hence given terms are in AP

It is said that, are in AP,

Hence,

Taking LCM,

.

Multiply in both denominator and numerator with c in LHS and a in RHS

Since,

a(b + c), b(c + a), c(a + b) are to be proved in A.P.

b(c + a)-a(b + c) = c(a + b)-b(c + a)

bc + ba – ab - ca- = ca + ab – bc - ba

cb - ca = ca - ab

c(b - a) = a(c - b)

Hence, given terms are in AP.

If a², b², c² are in A.P then, b² - a² = c² - b²

If are in A.P. then,

Since, b² - a² = c² - b²

Put b² - a²= c² - b² in above,

We get LHS = RHS

Hence given terms are in AP

(i)

b²c + b²a – a²b – a²c = c²a + c²b – b²a – b²c

c(b² - a² ) + ab(b - a) = a(c² - b² ) + bc(c - b)

(b - a)(ab + bc + ca) = (c - b)(ab + bc + ca)

b - a = c - b

And since a, b, c are in AP,

b - a = c - b

Hence given terms are in AP

(ii) b + c - a, c + a - b, a + b - c are in A.P.

Therefore, (c + a - b) - (b + c - a) = (a + b - c) - (c + a - b)

2a - 2b = 2b - 2c

b - a = c - b

And since a, b, c are in AP,

b - a = c - b

Hence given terms are in AP.

(iii) bc – a², ca – b², ab – c² are in A.P.

(ca - b²) – (bc - a²) = (ab - c²) – (ca - b²)

(a - b)(a + b + c) = (b - c)(a + b + c)

a - b = b - c

b - a = c - b

And since a, b, c are in AP,

b - a = c - b

Hence given terms are in AP

Since, are in AP,

a(b - c) = c(a - b)

Since, are in A.P.

Multiply in both denominator and numerator with,

C in LHS and a in RHS

Hence given terms are in AP

Since, bc, ca, ab are in A.P.

=

=

Since, are in A.P.

Multiply in both denominator and numerator with,

C in LHS and a in RHS

Hence given terms are in AP

(a - c)² = 4 (a - b) (b - c)

a² + c² - 2ac = 4(ab – ac – b² + bc)

a² + 4c²b² + 2ac - 4ab - 4bc = 0

(a + c - 2b)² = 0

a + c - 2b = 0

Since a, b, c are in AP

b - a = c - b

a + c - 2b = 0

Hence,

(a - c)² = 4 (a - b) (b - c)

a² + c² + 4ac = 2 (ab + bc + ca)

a² + c² + 2ac - 2ab - 2bc = 0

(a + c - b)² – b² = 0

a + c - b = b

a + c - 2b = 0

Since a, b, c are in AP

b - a = c - b

a + c - 2b = 0

Hence proved

a³ + c³ + 6abc = 8b³

a³ + c³ - (2b)³ + 6abc = 0

a³ + (-2b)³ + c³ + 3a(-2b)c = 0

Since, if a + b + c = 0, a³ + b³ + c³ = 3abc

(a - 2b + c)³ = 0

a - 2b + c = 0

Since a, b, c are in AP

b - a = c - b

= a + c - 2b = 0

Hence proved

Since are in A.P

Also are also in AP

Therefore,

= are in AP

= are in AP

Multiply by abc in numerator in all terms,

= are in AP

= a, b, c are in AP

Hence proved

Since, x² + xy + y², z² + zx + x² and y² + yz + z² are in AP

(z² + zx + x²) - (x² + xy + y²) = (y² + yz + z²) - (z² + zx + x²)

Let d = common difference,

Sicex, y, z are in AP

Y = x + d and x = x + 2d

Therefore the above equation becomes,

= (x + 2d)² + (x + 2d)x - x(x + d) - (x + d)² = (x + d)² + (x + d)(x + 2d) - (x + 2d)x – x²

= x² + 4xd + 4d² + x² + 2xd – x² – xd – x² - 2xd – d² = x² + 2dx + d² + x² + 2dx + xd + 2d² – x² - 2dx – x²

Hence, proved.

x² + xy + y², z² + zx + x² and y² + yz + z² are in consecutive terms of an A.P.

(i) Let A be the Arithmetic mean

Then 7, A, 13 are in AP

A-7 = 13-A

2A = 13 + 7

A = 10

(ii) Let A be the Arithmetic mean

Then 12, A, - 8 are in AP

A - 12 = - 8 - A

2A = 12 + 8

A = 2

(iii) Let A be the Arithmetic mean

Then x - y, A, x + y are in AP

A - (x - y) = (x + y) – A

2A = x + y + x - y

A = x

Let A₁, A₂, A₃, A₄ be the 4 AM Between 4 And 19

Then, 4, A₁, A₂, A₃, A₄, 19 are in AP.

We know,

d = 3

Hence,

A₁ = a + d = 4 + 3 = 7

A₂ = A1 + d = 7 + 3 = 10

A₃ = A2 + d = 10 + 3 = 13

A₄ = A3 + d = 13 + 3 = 16

Let A₁, A₂, A₃, A₄, A₅, A₆, A₇ be the 7 AMs Between 2 And 17

Then, 2, A₁, A₂, A₃, A₄, A₅, A₆, A₇, 17 are in AP

We know,

A_n = a + (n - 1)d

A₉ = 17 = 2 + (9 - 1)d

Therefore,

So, 7 AMs between 2 and 7 are

Let A₁, A₂, A₃, A₄, A₅, A₆ be the 7 AM Between 15 And - 13

Then, 15, A₁, A₂, A₃, A₄, A₅, A₆, - 13 are in AP

We know,

A_n = a + (n - 1)d

A_{8 =} - 13 = 15 + (8 - 1)d

d = -4

Hence,

A₁ = a + d = 15 - 4 = 11

A₂ = A1 + d = 11 - 4 = 7

A₃ = A2 + d = 7 - 4 = 3

A₄ = A3 + d = 3 - 4 = -1

A₅ = A4 + d = -1 - 4 = -5

A₆ = A5 + d = -5 - 4 = -9

Let the series be 3, A₁, A₂, A₃, ........, A_n, 17

From the data,

We know total terms in AP are n + 2

So 17 is the (n + 2)th term

We know,

A_n = a + (n - 1)d

So, 17 = 3 + (n + 2 - 1)d

Therefore,

And

Since,

9n + 51 = 17n + 3

8n = 48

n = 6

There are 6 terms in AP

Let the series be 7, A₁, A₂, A₃, ........, A_n, 71

We know total terms in AP are n + 2

So 71 is the (n + 2)th term

We know,

A_n = a + (n - 1)d

So, A₆ = a + (6 - 1)d

a + 5d = 27 (5^th term)

d = 4

71 = (n + 2)th term

71 = a + (n + 2 - 1)d

71 = 7 + n(4)

n = 15

There are 15 terms in AP

Let a and b be the first and last terms and

The series be a, A₁, A₂, A₃, ........, A_n, b

So We know, Mean

Mean of A₁ and A_n =

A₁ = a + d

A_n = a - d

Therefore, AM

AM between A₂ and A_n-1

Similarly, it is (a + b)/2 for all such numbers, which is constant

Hence, AM = (a + b)/2

Given that,

A₁ = AM of x and y

And A₂ = AM of y and z

So,

AM of A₁ and A₂

Since x, y, z are in AP,

Finally, AM =

Hence proved.

Let A₁, A₂, A₃, A₄, A₅ be the 5 nos Between 8 And 26

Then, 8, A₁, A₂, A₃, A₄, A₅, 26 are in AP

We know,

A_n = a + (n - 1)d

A_{7 =} 26 = 8 + (7 - 1)d

d = 3

Hence,

A₁ = a + d = 8 + 3 = 11

A₂ = A₁ + d = 11 + 3 = 14

A₃ = A₂ + d = 14 + 3 = 17

A₄ = A₃ + d = 17 + 3 = 20

A₅ = A₄ + d = 20 + 3 = 23

Given: A man saved 16500/- in ten years

To find: His saving in the first year

Let he saved Rs. x in the first year

Since each year after the first he saved 100/- more then he did in the preceding year

So,

A.P will be x, 100 + x, 200 + x………………..

where x is first term and

common difference, d = 100 + x – x = 100

We know,

S_n is the sum of n terms of an A.P

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

According to the question:

S_n = 16500 and n = 10

Therefore,

⇒ 16500 = 5{2x + 9(100)}

⇒ 16500 = 5(2x + 900)

⇒ 16500 = 10x + 4500

⇒ -10x = 4500 – 16500

⇒ –10x = –12000

⇒ x = 1200

Hence, his saving in first year is 1200

Given: A man saves Rs. 32 during first year and increases his savings by Rs. 4 every year. His total saving is Rs. 200

To find: Time taken in years by him to save Rs. 200.

A man saves in the first year is 32Rs

He saves in the second year is 36Rs.

In this process he increases his savings by Rs. 4 every year

Therefore,

A.P. will be 32, 36, 40,…………………

where 32 is first term and

common difference, d = 36 – 32 = 4

We know,

S_n is the sum of n terms of an A.P

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

According to the question:

S_n = 200

Therefore,

n can not be a negative values as days can not hold negative values

⇒ n = 5

Hence, he has to do saving for 5 days to save Rs. 200

Given: Total debt is Rs. 3600 and total number of installments are 40 and in 30 installments, he has paid two-third of the debt and dies leaving one-third of amount

Let first instalment be “a”

Let the common difference of the instalments be “d”

Here, S₄₀ = 3600 and n = 40

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ 2a+39d = 180 ………(1)

Since 30 installments are paid and one third of the debt unpaid

So,

S₃₀ = 2400 and n = 30

⇒ 2a+29d = 160 ………(2)

Solving (1) and(2) by substitution method,

2a + 39d = 180

⇒ 2a = 180 – 39d ……(3)

Put value of 2a from (3) in (2):

2a + 29d = 160

⇒ 180 – 39d + 29d = 160

⇒ –10d = 160 – 180

⇒ –10d = –20

⇒ d = 2

Put this value of d in (3):

2a = 180 – 39d

⇒ 2a = 180 – 39(2)

⇒ 2a = 180 – 78

⇒ 2a = 102

⇒ a = 51

Hence, value of first installment = a = Rs. 51

To find: the sum of totals products in 7 years i.e. S₇

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ S₇ = 7(625)

⇒ S₇ = 4375

Hence, the total product in the 7 years are 4375

(ііі) the product in the 10^th year

Answer:

To find: the product in the 10^th year i.e. a₁₀

Formula used:

For an A.P., a_n is n^th term which is given by,

a_n = a + (n – 1)d

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

a₁₀ = 550 + (10 – 1)25

⇒ a₁₀ = 550 + (9)25

⇒ a₁₀ = 550 + 225

⇒ a₁₀ = 775

Hence, the product in the 10^th year are 775 units

Given: total trees are 25 and equal distance between two adjacent trees are 5 meters

To find: the total distance the gardener will cover

As gardener is coming back to well after watering every tree

Distance covered by him to water 1^st tree and return to the initial position is 10m + 10m = 20m

Now, distance between adjacent trees is 5m

Distance covered by him to water 2^nd tree and return to the initial position is 15m + 15m = 30m

Distance covered by the gardener to water 3^rd tree return to the initial postion is 20m + 20m = 40m

Hence distance covered by the gardener to water the trees are in A.P

A.P. is 20, 30, 40 ………upto 25 terms

Here first term,a = 20,common difference, d = 30 – 20 = 10

And n = 25

We need to find S₂₅ which will be the total distance covered by gardener to water 25 trees

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ S₂₅ = 25(140)

⇒ S₂₅ = 3500

Hence, total distance covered by gardener to water trees is 3500 m

Given: Amount to be counted is Rs. 10710

To find: Time taken by man to count the entire amount

He counts at the rate of Rs. 180 per minute for half an hour or 30 minutes

Amount to be counted in an hour = 180 * 30 = Rs. 5400

Amount to be left = 10710 – 5400 = 5310

⇒ S_n = 5310

After an hour, rate of counting is decreasing at Rs. 3 per minute. This rate will form an A.P.

A.P. will be 177, 174, 171,……………………

Here a = 177 and d = 174 – 177 = –3

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ 10620 = 3n(119 – n)

⇒ n² – 59n – 60n + 3540 = 0

⇒ n(n – 59) – 60(n – 59) = 0

⇒ (n – 59)(n – 60) = 0

⇒ n = 59 or 60

We will take value of n = 59 as at 60^th min he will count Rs.0

Therefore, total time taken by him to count the entire amount = 30 + 59 = 89 minutes

Given: A piece of equipment cost a certain factory is 600,000

To find: Value of the equipment at the end of 10 years

It depreciates 15%, 13.5%, 12% in 1^st, 2^nd, 3^rd year and so on.

This means price of equipment is depreciating in an A.P.

A.P. will be 15, 13.5, 12,…………………………upto 10 terms

Hence a = 15, d = 13.5 – 15 = –1.5

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

Total percentage of depreciation in 10 years,

⇒ S₁₀ = 82.5

Value of the equipment at the end of 10 years,

= 175 × 600

= 105000

Hence, value of equipment at the end of 10 years is Rs. 105000

Given: Price of the tractor is Rs. 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount.

To find: Total cost of the tractor if he buys it at installment

Total price = 12000

Paid amount = 6000

Unpaid amount = 12000 – 6000 = 6000

He pays remaining 6000 in n number of installments of 500 each

Cost incurred by him to pay remaining 6000,

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

= 10680

Total cost = 6000 + 10680 = 16680

Hence, total cost of the tractor if he buys it at installment is Rs. 16680

Given: Price of the scooter is Rs. 22000. He pays ₹ 4000 cash and agrees to pay the balance in annual instalments of Rs. 1000 plus 10% interest on the unpaid amount.

To find: Total cost of the scooter if he buys it at installment

Total price = 22000

Paid amount = 4000

Unpaid amount = 22000 – 4000 = 18000

He pays remaining 18000 in n number of installments of 1000 each

Cost incurred by him to pay remaining 18000,

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

= 35100

Total cost = 4000 + 35100 = 39100

Hence, total cost of the scooter if he buys it at installment is Rs. 39100

Given: The income of a person is ₹ 300, 000 in the first year

and he receives an increase of ₹ 10, 000 to his income per year for the next 19 years.

To find: the total amount he received in 20 years i.e. S₂₀

According to question:

a = 300000, d = 10000 and n = 20

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ S₂₀ = 10(790000)

⇒ S₂₀ = 7900000

Hence, the total amount he received at the end of 20 years is Rs. 790000

Given: First installment is Rs. 100 and he increases it by Rs. 5 every month

To find: Total amount in 30^th installment i.e. a₃₀

According to question:

A.P. will be 100, 105,……………………

a = 100 and d = 5

Formula used:

a_n = a + (n – 1)d

⇒ a₃₀ = 100 + (30 – 1)5

⇒ a₃₀ = 100 + (29)5

⇒ a₃₀ = 100 + 145

⇒ a₃₀ = 245

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

Given: Carpenter was hired to build 192 frames. He made 5 frames on first day and two more frames for every next day

To find: Number of days taken by him to build 192 frames

According to question:

A.P. will be 5, 7, 9,………………………………

⇒ S_n = 192, a = 5 and d = 7 – 5 = 2

Formula used:

where a is first term, d is common difference and n is number of terms in an A.P.

Therefore,

⇒ n² + 16n – 12n + 192 = 0

⇒ n(n + 16) – 12(n + 16) = 0

⇒ (n + 16)(n – 12) = 0

⇒ n = –16 or 12

n is number of days and it cannot be negative

Therefore, we will take value of n = 12

Hence, he will take 12 days to build 192 frames

Given: the sum of interior angles of a triangle is 180°

To prove: Sum of interior angles of polygons with 3, 4, 5,…… forms an A.P.

To find: Sum of interior angles for a 21 sided polygon i.e. a₂₁

We know,

The sum of interior angles of a polygon with n sides,

a_n = 180° × (n – 2)

Sum of interior angles of a polygon of 3 sides,

a₃= 180° × (3 – 2) = 180° × 1 = 180°

Sum of interior angles of a polygon of 4 sides,

a₄= 180° × (4 – 2) = 180° × 2 = 360°

Sum of interior angles of a polygon of 5 sides,

a₅= 180° × (5 – 2) = 180° × 3 = 540°

A.P is known for Arithmetic Progression whose common difference, d = a_n – a_n-1 where n > 0

Here,

d = a₄ – a₃ = 360° - 180° = 180°

d = a₅ – a₄ = 540° - 360° = 180°

Common difference is same in both the cases

This shows that sum of interior angles of polygons with 3, 4, 5,…… forms an A.P.

Hence Proved

Sum of interior angles of a polygon of 21 sides,

a₂₁= 180° × (21 – 2) = 180° × 19 = 3420°

Given: total potatoes are 20. First potato is 24 meters away from starting point equal distance between two adjacent potatoes is 4 meters

To find: the total distance run by contestant

As contestant is coming back to starting point after bringing every potato

Distance covered by him to bring 1^st potato and return to the initial position is 24m + 24m = 48m

Now, distance between adjacent trees is 4m

Distance covered by him to bring 2^nd potato and return to the initial position is 28m + 28m = 56m

Distance covered by him to bring 3^rd potato return to the initial postion is 32m + 32m = 64m

Hence distance covered by him to bring the potatoes are in A.P

A.P. is 48, 56, 64………………………………………upto 20 terms

Here first term, a = 48, common difference, d = 56 – 48 = 8

And n = 20

We need to find S₂₀ which will be the total distance covered by contestant to bring 20 potatoes