Question

Find the sum of first 17 terms of an AP whose 4^th and 9^th terms are –15 and –30 respectively.

Answer 1

Let the first term, common difference and the number of terms in an AP are a, d and n, respectively.

We know that, the n^th term of an AP,

T_n = a + (n – 1)d  ...(i)

∴ 4^th term of an AP,

T₄ = a + (4 – 1)d = –15   ...[Given]

⇒ a + 3d = –15  ...(ii)

And 9^th term of an AP,

T₉ = a + (9 – 1)d = –30 ...[Given]

⇒ a + 8d = –30   ...(iii)

Now, subtract equation (ii) from equation (iii), we get

a + 8d = –30
a + 3d = –15
– –         +      
      5d = –15

⇒ d = –3

Put the value of d in equation (ii), we get

a + 3(–3) = –15

⇒ a – 9 = –15

⇒ a = –15 + 9 = – 6

∵ Sum of first n terms of an AP,

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

∴ Sum of first 17 terms of an AP,

S₁₇ = `17/2 [2 xx (-6) + (17 - 1)(-3)]`

= `17/2 [-12 + (16)(-3)]`

= `17/2(-12 - 48)`

= `17/2 xx (-60)`

= 17 × (–30)

= –510

Hence, the required sum of first 17 terms of an AP is –510.