Question

Find the sum of first $17$ terms of an $A.P.$, where $4th$ and $9th$ terms are $-15$ and $-30$, respectively.

Answer 1

$T_4=-15=a+(4-1)d$

$-15=a+\left(3\right)d$ (1)

$T_9=-30=a+(9-1)d$

$-30=a+\left(8\right)d$ (2)

Where a is the first term of A.P and d is the difference between two terms.

(2)-(1) gives

$-15=5d$

$d=-3$ (3)

Substiyutin (3) in (1)

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

$-15=a-9$

$a=-6$

$S_{17}=\frac{n}{2}(2a+(n-1\left)d\right)$

$S_{17}=\frac{17}{2}\left(2\right(-6)+(17-1\left)d\right)$

$S_{17}=\frac{17}{2}(-12+(16\left)\right(-3\left)\right)$

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

$S_{17}=17\times -30=-510$

Hence the correct answer is $-510$