Find the sum of first 17 terms of an AP whose \( 4^{\text {th }} \) and \( 9^{\text {th }} \) terms are \( -15 \) and \( -30 \) respectively.

Given: 

The \( 4^{\text {th }} \) and \( 9^{\text {th }} \) terms of an AP are \( -15 \) and \( -30 \) respectively.

To do: 

We have to find the sum of first 17 terms.

Solution:

Let $a$ be the first term and $d$ be the common difference.

This implies,

$a_{4}=a+(4-1)d$

$-15=a+3d$

$a=-15-3d$.........(i)

$a_9=a+(9-1)d$

$-30=a+8d$

$-30=(-15-3d)+8d$           [From (i)]

$-30=-15-3d+8d$

$-30+15=5d$

$5d=-15$

$d=-3$

This implies,

$a=-15-3(-3)$

$=-15+9$

$=-6$

We know that,

$S_n=\frac{n}{2}[2a+(n-1)d$

$S_{17}=\frac{17}{2}[2(-6)+(17-1)(-3)]$

$=\frac{17}{2}[-12-16(3)]$

$=\frac{17}{2}(-12-48)$

$=\frac{17}{2}(-60)$

$=17(-30)$

$=-510$

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Updated on: 10-Oct-2022

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