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8. Sequences and Series
hard
Let $S_n$ denote the sum of the first $n$ terms of an $A.P$.. If $S_4 = 16$ and $S_6 = -48$, then $S_{10}$ is equal to
A
$-410$
B
$-260$
C
$-320$
D
$-380$
(JEE MAIN-2019)
Solution
$2\left\{ {2a + 3d} \right\} = 16$
$3\left\{ {2a + 5d} \right\} = – 48$
$2a + 3d = 8$
$2a + 5d = – 16$
$d = – 12$
${S_{10}} = 5\left\{ {44 – 9 \times 12} \right\}$
$ = – 320$
Standard 11
Mathematics