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
Show Answer
Answer:
Correct Answer: 2. (c)
Solution:
- 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, $\quad S _{10}=5[2 a+9 d]$
$$ =5[44+9(-12)]=5[44-108] $$
$$ =5 \times(-64)=-320 $$
