Question

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.

• A. -1 , -1 , 3 , 5 ...
• B. 1 , 1 , 3 , -5 ...
• C. 1 , -1 , -3 , -5 ...
• D. -1 , -1 , -3 , 5 ...

Answer 1

The correct option is C

1 , -1 , -3 , -5 ...

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

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

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

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

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

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

Sum of the first 20 terms = -80 -280 = -360

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

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

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

Subtracting (i) from (ii) , we have
$10d=-20\Rightarrow d=-2.$

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

Hence the required AP is
$a,a+d,a+2d,a+3d...=1,1-2,1-4,1-\mathrm{6...}=1,-1,-3,-5...$