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

(c) : Let a and d be the first term and common difference of the A.P. respectively. Given S_4=16 => (4 / 2)(2 a+3 d)=16 => 2 a+3 d=8 (i) Also, S_6=-48 => (6 / 2)(2 a+5 d)=-48 => 2 a+5 d=-16 (ii) Solving (i) and (ii), we get d=-12, a=22 Now, S_10=(10 / 2)[2(22)+9(-12)]=5(44-108)=-320

Let S_{n} denote the sum of the first n terms of an A.P... | Sequence & Series (Mathematics)

(c) : Let $a$ and $d$ be the first term and common $\\$ difference of the A.P. respectively. $\\$ Given $S_4=16 \Rightarrow \frac{4}{2}(2 a+3 d)=16 \Rightarrow 2 a+3 d=8$ (i) $\\$ Also, $S_6=-48$ $\Rightarrow \frac{6}{2}(2 a+5 d)=-48 \Rightarrow 2 a+5 d=-16$ (ii) $\\$ Solving (i) and (ii), we get $d=-12, a=22\\$ Now, $S_{10}=\frac{10}{2}[2(22)+9(-12)]=5(44-108)=-320$

Explanation

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