Question

If the sum of first 7 terms of an A.P. is 49 and that of its first 17 terms is 289, find the sum of first n terms of the A.P.

The sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first n terms.

If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.

Answer 1

Let the first term and the common difference of the given A.P. be a and d, respectively.

Sum of the first 7 terms, S₇ = 49

We know

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

⇒ `7/2(2a + 6d) = 49`

⇒ `7/2 xx 2(a + 3d) = 49`

⇒ a + 3d = 7   ...(1)

Sum of the first 17 terms, S₁₇ = 289

⇒ `17/2(2a + 16d) = 289`

⇒ `17/2 xx 2(a + 8d) = 289`

⇒ a + 8d = `289/17`

⇒ a + 8d = 17   ...(2)

Subtracting (2) from (1), we get

5d = 10

d = `5/10`

⇒ d = 2

Substituting the value of d in (1), we get

a = 1

Now,

sum of the first n terms is given by

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

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

= `n/2 [2 + 2n - 2]`

= `n/2 [2n]`

= n²

Therefore, the sum of the first n terms of the A.P. is n².