Sequences and Series 2 Question 2

2. Let $S _n$ denote the sum of the first $n$ terms of an AP. If $S _4=16$ and $S _6=-48$, then $S _{10}$ is equal to

(2019 Main, 12 April I)

(a) -260

(b) -410

(c) -320

(d) -380

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Answer:

Correct Answer: 2. (c)

Solution:

  1. Given $S _n$ denote the sum of the first $n$ terms of an AP.

Let first term and common difference of the AP be ’ $\alpha$ ’ and ’ $d$ ‘, respectively.

$$ \begin{aligned} & \therefore \quad S _4=2[2 a+3 d]=16 \\ & \because S _n=\frac{n}{2}[2 a+(n-1) d] \\ & \Rightarrow \quad 2 a+3 d=8 \quad \ldots \text { (i) } \\ & \text { and } \quad S _6=3[2 a+5 d]=-48 \quad \text { [given] } \\ & \Rightarrow \quad 2 a+5 d=-16 \end{aligned} $$

On subtracting Eq. (i) from Eq. (ii), we get

$$ \begin{array}{rlrl} & & 2 d & =-24 \\ \Rightarrow & d & =-12 \end{array} $$

So, $\quad 2 a=44 \quad$ [put $d=-12$ in Eq. (i)]

Now, $\quad S _{10}=5[2 a+9 d]$

$$ =5[44+9(-12)]=5[44-108] $$

$$ =5 \times(-64)=-320 $$



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