The sum of first 10 terms of an AP is −150 and the sum of its next 10 terms is −550.

Solution:

Let a be the first term and d be the common difference of the AP. Then,

$S_{10}=-150$                                     (Given)

$\Rightarrow \frac{10}{2}(2 a+9 d)=-150 \quad\left\{S_{n}=\frac{n}{2}[2 a+(n-1) d]\right\}$

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

$\Rightarrow 2 a+9 d=-30 \quad \ldots(1)$

It is given that the sum of its next 10 terms is −550.

Now,

S20 = Sum of first 20 terms = Sum of first 10 terms + Sum of the next 10 terms = −150 + (−550) = −700

$\therefore S_{20}=-700$

$\Rightarrow \frac{20}{2}(2 a+19 d)=-700$

$\Rightarrow 10(2 a+19 d)=-700$

$\Rightarrow 2 a+19 d=-70 \quad \ldots .(2)$

Subtracting (1) from (2), we get

$(2 a+19 d)-(2 a+9 d)=-70-(-30)$

$\Rightarrow 10 d=-40$

$\Rightarrow d=-4$

Putting d = −4 in (1), we get

$2 a+9 \times(-4)=-30$

$2 a+9 \times(-4)=-30$

$\Rightarrow 2 a=-30+36=6$

$\Rightarrow a=3$

Hence, the required AP is 3, −1, −5, −9, ... .