In an AP, the sum of the first ten terms is $-80$ and the sum of the next ten terms is $-280$. Find the AP.

The correct option is A $1,-1,-3,-5\dots$

Sum of the first 10 terms of the AP $=-80$

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

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

$\Rightarrow -80=5(2a+9d)$

$\Rightarrow -16=2a+9d\cdots \left(i\right)$

Sum of the next $10$ terms of the AP $=-280$

Sum of the first $20$ terms = Sum of the first 10 terms of the AP + Sum of the next $10$ terms of the AP $=-80+(-280)=-80-280=-360$

i..e, $S_{20}=\frac{20}{2}(2a+(20-1\left)d\right)=-360$

$\Rightarrow -360=10(2a+19d)$

$\Rightarrow -36=2a+19d\cdots \left(ii\right)$

Subtracting $\left(i\right)$ from $\left(ii\right)$ , we have

$2a+19d-(2a+9d)=-36-(-16)$

$\Rightarrow 10d=-20$

$\Rightarrow d=-2$

Subsituting this value of $d$ in $\left(i\right)$, we have
$-16=2a-18.$
$\Rightarrow a=1$

Hence the required AP is
$a,a+d,a+2d,a+3d\dots$
$=1,(1-2),(1-4),(1-6)\dots$
$=1,-1,-3,-5\dots$