Question:
Let $S_{n}$ denote the sum of the first $n$ terms of an A.P. If $\mathrm{S}_{4}=16$ and $\mathrm{S}_{6}=-48$, then $\mathrm{S}_{10}$ is equal to:
Correct Option: 1
Solution:
$2\{2 \mathrm{a}+3 \mathrm{~d}\}=16$
$3(2 \mathrm{a}+5 \mathrm{~d})=-48$
$2 \mathrm{a}+3 \mathrm{~d}=8$
$2 \mathrm{a}+5 \mathrm{~d}=-16$
$\mathrm{d}=-12$
$\mathrm{S}_{10}=5\{44-9 \times 12\}$
$=-320$
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