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

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