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