Here, a = 9, d = 17 - 9 = 8 and S_n = 636. The sum of n terms of an AP is given by S_n = n/2[2a + (n -1)d] ⇒ 636 = n/2[2(9) + (n -1)(8)] ⇒ 636 = n[9 + 4n - 4] ⇒ 4n² + 5n - 636 = 0 ⇒ 4n² + 53n - 48n - 636 = 0 ⇒ n(4n + 53) - 12(4n + 53) = 0 ⇒ (n - 12)(4n + 53) = 0 ⇒ n - 12 = 0 [∵ sn + 53 ≠ 0 as n ≠ -Read more

Here, a = 9, d = 17 – 9 = 8 and S_n = 636.
The sum of n terms of an AP is given by
S_n = n/2[2a + (n -1)d]
⇒ 636 = n/2[2(9) + (n -1)(8)]
⇒ 636 = n[9 + 4n – 4]
⇒ 4n² + 5n – 636 = 0
⇒ 4n² + 53n – 48n – 636 = 0
⇒ n(4n + 53) – 12(4n + 53) = 0
⇒ (n – 12)(4n + 53) = 0
⇒ n – 12 = 0 [∵ sn + 53 ≠ 0 as n ≠ – 53/4]
⇒ n = 12
Hence, 12 terms of the AP: 9, 17, 25 … must be taken to get the sum 636.

Here, a = 5, a_n = 45 and S_n = 400. The sum of n terms of an AP is given by S_n = n/2[a + a_n] ⇒ 400 = n/2[5 + 45] ⇒ 400 = 25n ⇒ n = 400/25 = 16 a_n = a + (n - 1)d ⇒ 45 = 5 + (16 - 1)d ⇒ 40 = 15d ⇒ d = 40/15 = 8/3 Hence, the number of term are 16 and the common difference is 8/3.

Here, a = 5, a_n = 45 and S_n = 400.
The sum of n terms of an AP is given by
S_n = n/2[a + a_n]
⇒ 400 = n/2[5 + 45]
⇒ 400 = 25n
⇒ n = 400/25 = 16
a_n = a + (n – 1)d
⇒ 45 = 5 + (16 – 1)d
⇒ 40 = 15d
⇒ d = 40/15 = 8/3
Hence, the number of term are 16 and the common difference is 8/3.

Here, a = 17, a_n = 350 and d = 9. a_n = a + (n - 1)d ⇒ 350 = 17 + (n -1)9 ⇒ 350 = 8 + 9n ⇒ n = 342/9 = 38 The sum of n terms of an AP is given by S_n = n/2[a + a_n] ⇒ S₃₈ = 38/2[17 + 350] ⇒ S₃₈ = 19 × 367 = 6973 Hence, there are 38 terms and their sum is 6973.

Here, a = 17, a_n = 350 and d = 9.
a_n = a + (n – 1)d
⇒ 350 = 17 + (n -1)9
⇒ 350 = 8 + 9n
⇒ n = 342/9 = 38
The sum of n terms of an AP is given by
S_n = n/2[a + a_n]
⇒ S₃₈ = 38/2[17 + 350]
⇒ S₃₈ = 19 × 367
= 6973
Hence, there are 38 terms and their sum is 6973.

Here, a = 34 and d = 32 - 34 = - 2. Let, the nth term of the A.P. is 10. Therefore, a_n = 10 ⇒ a + (n -1)d = 10 ⇒ 34 + (n - 1)(-2) = 10 ⇒ (n -1)(- 2) = - 24 ⇒ n -1 = 12 ⇒ n = 13 The sum of n terms of an AP is given by S_n = n/2[a + l] ⇒ S₁₃ = 13/2[34 + 10] ⇒ S₁₃ = 13/2[44] = 13 × 22 = 286

Here, a = 34 and d = 32 – 34 = – 2.
Let, the nth term of the A.P. is 10.
Therefore, a_n = 10
⇒ a + (n -1)d = 10
⇒ 34 + (n – 1)(-2) = 10
⇒ (n -1)(- 2) = – 24
⇒ n -1 = 12
⇒ n = 13
The sum of n terms of an AP is given by
S_n = n/2[a + l]
⇒ S₁₃ = 13/2[34 + 10]

A.P.: 2, 7, 12, ... Here, a = 2 and d = 7 - 2 = 5 The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] ⇒ S_10 = 10/2[2(2) + (10 - 1)(5)] ⇒ S_10 = 5[4 + 45] = 245.

A.P.: 2, 7, 12, …
Here, a = 2 and d = 7 – 2 = 5
The sum of n terms of an AP is Given by
Sn = n/2[2a + (n -1)d]
⇒ S_10 = 10/2[2(2) + (10 – 1)(5)]
⇒ S_10 = 5[4 + 45] = 245.

## How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Here, a = 9, d = 17 - 9 = 8 and S_n = 636. The sum of n terms of an AP is given by S_n = n/2[2a + (n -1)d] ⇒ 636 = n/2[2(9) + (n -1)(8)] ⇒ 636 = n[9 + 4n - 4] ⇒ 4n² + 5n - 636 = 0 ⇒ 4n² + 53n - 48n - 636 = 0 ⇒ n(4n + 53) - 12(4n + 53) = 0 ⇒ (n - 12)(4n + 53) = 0 ⇒ n - 12 = 0 [∵ sn + 53 ≠ 0 as n ≠ -Read more

Here, a = 9, d = 17 – 9 = 8 and S_n = 636.

See lessThe sum of n terms of an AP is given by

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

⇒ 636 = n/2[2(9) + (n -1)(8)]

⇒ 636 = n[9 + 4n – 4]

⇒ 4n² + 5n – 636 = 0

⇒ 4n² + 53n – 48n – 636 = 0

⇒ n(4n + 53) – 12(4n + 53) = 0

⇒ (n – 12)(4n + 53) = 0

⇒ n – 12 = 0 [∵ sn + 53 ≠ 0 as n ≠ – 53/4]

⇒ n = 12

Hence, 12 terms of the AP: 9, 17, 25 … must be taken to get the sum 636.

## The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Here, a = 5, a_n = 45 and S_n = 400. The sum of n terms of an AP is given by S_n = n/2[a + a_n] ⇒ 400 = n/2[5 + 45] ⇒ 400 = 25n ⇒ n = 400/25 = 16 a_n = a + (n - 1)d ⇒ 45 = 5 + (16 - 1)d ⇒ 40 = 15d ⇒ d = 40/15 = 8/3 Hence, the number of term are 16 and the common difference is 8/3.

Here, a = 5, a_n = 45 and S_n = 400.

See lessThe sum of n terms of an AP is given by

S_n = n/2[a + a_n]

⇒ 400 = n/2[5 + 45]

⇒ 400 = 25n

⇒ n = 400/25 = 16

a_n = a + (n – 1)d

⇒ 45 = 5 + (16 – 1)d

⇒ 40 = 15d

⇒ d = 40/15 = 8/3

Hence, the number of term are 16 and the common difference is 8/3.

## The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Here, a = 17, a_n = 350 and d = 9. a_n = a + (n - 1)d ⇒ 350 = 17 + (n -1)9 ⇒ 350 = 8 + 9n ⇒ n = 342/9 = 38 The sum of n terms of an AP is given by S_n = n/2[a + a_n] ⇒ S₃₈ = 38/2[17 + 350] ⇒ S₃₈ = 19 × 367 = 6973 Hence, there are 38 terms and their sum is 6973.

Here, a = 17, a_n = 350 and d = 9.

See lessa_n = a + (n – 1)d

⇒ 350 = 17 + (n -1)9

⇒ 350 = 8 + 9n

⇒ n = 342/9 = 38

The sum of n terms of an AP is given by

S_n = n/2[a + a_n]

⇒ S₃₈ = 38/2[17 + 350]

⇒ S₃₈ = 19 × 367

= 6973

Hence, there are 38 terms and their sum is 6973.

## Find the sums given below :34 + 32 + 30 + . . . + 10

Here, a = 34 and d = 32 - 34 = - 2. Let, the nth term of the A.P. is 10. Therefore, a_n = 10 ⇒ a + (n -1)d = 10 ⇒ 34 + (n - 1)(-2) = 10 ⇒ (n -1)(- 2) = - 24 ⇒ n -1 = 12 ⇒ n = 13 The sum of n terms of an AP is given by S_n = n/2[a + l] ⇒ S₁₃ = 13/2[34 + 10] ⇒ S₁₃ = 13/2[44] = 13 × 22 = 286

Here, a = 34 and d = 32 – 34 = – 2.

Let, the nth term of the A.P. is 10.

Therefore, a_n = 10

⇒ a + (n -1)d = 10

⇒ 34 + (n – 1)(-2) = 10

⇒ (n -1)(- 2) = – 24

⇒ n -1 = 12

⇒ n = 13

The sum of n terms of an AP is given by

S_n = n/2[a + l]

⇒ S₁₃ = 13/2[34 + 10]

⇒ S₁₃ = 13/2[44]

See less= 13 × 22 = 286

## Find the sum of the following APs: 2, 7, 12, . . ., to 10 terms.

A.P.: 2, 7, 12, ... Here, a = 2 and d = 7 - 2 = 5 The sum of n terms of an AP is Given by Sn = n/2[2a + (n -1)d] ⇒ S_10 = 10/2[2(2) + (10 - 1)(5)] ⇒ S_10 = 5[4 + 45] = 245.

A.P.: 2, 7, 12, …

See lessHere, a = 2 and d = 7 – 2 = 5

The sum of n terms of an AP is Given by

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

⇒ S_10 = 10/2[2(2) + (10 – 1)(5)]

⇒ S_10 = 5[4 + 45] = 245.