If in an AP, the sum of first 10 terms is -150 and the sum of its next 10 terms is -550. Find the AP.

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Solution

Given,
$S_{10} = - 150$
$S_{20} - S_{10} =$ Next 10 term
$S_{20} - S_{10} = - 550$
$S_{20} = - 550 + S_{10}$
$S_{20} = - 550 - 150$
$S_{20} = - 700$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$S_{10} = \frac{10}{2} [2 a + (10 - 1) d] = - 150$

$2 a + 9 d = - 30$ (i)
$S_{20} = - 700$

$\frac{20}{2} [2 a + 19 d] = - 700$

$2 a + 19 d = - 70$ (ii)
eq(i)-eq(ii)
$(2 a + 9 d = - 30) - (2 a + 19 d = - 70)$
$- 10 d = 40$
$d = - 4$
$2 a + 9 d = - 30$
$2 a = - 30 + 36$
$\Rightarrow a = 3$
So the A.P. will be as:-
a, (a+d), (a+2d), ...
3, (-1), (-5) & so on

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