Question

In an $AP$, the sum of first ten terms is $-150$ and the sum of its next ten terms is $-550$. Find the $AP.$

Answer 1

$S_{10}=-150$

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

$S_{10}=\frac{10}{2}[2a+(10-1\left)d\right]$

$-150=5(2a+9d)$

$\therefore -30=2a+9d$...(1)

$-550=\frac{n}{2}\left[2\right(a+10d)+(n-1\left)d\right]$

$-550=\frac{10}{2}\left[2\right(a+10d)+(10-1\left)d\right]$

$-550=\frac{10}{2}[2a+20d+9d]$

$-550=5[2a+29d]$

$-110=2a+29d$...(2)

Solving (1) and (2) we get:

$d=-4,a=3$

Series: $\{3,-1,-5,-9,,,,,,,,,\}$