Sequences and Series 3 Question 1
1. Let $a, b$ and $c$ be in GP with common ratio $r$, where $a \neq 0$ and $0<r \leq \frac{1}{2}$. If $3 a, 7 b$ and $15 c$ are the first three terms of an AP, then the 4 th term of this AP is
(2019 Main, 10 April II)
(a) $5 a$
(b) $\frac{2}{3} a$
(c) $a$
(d) $\frac{7}{3} a$
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Answer:
Correct Answer: 1. (c)
Solution:
- Key Idea Use $n^{\text {th }}$ term of AP i.e., $a _n=a+(n-1) d$, If $a, A$, $b$ are in AP, then $2 A=a+b$ and $n^{\text {th }}$ term of G.P. i.e., $a _n=a r^{n-1}$.
It is given that, the terms $a, b, c$ are in GP with common ratio $r$, where $a \neq 0$ and $0<r \leq \frac{1}{2}$.
So, let, $b=a r$ and $c=a r^{2}$
Now, the terms $3 a, 7 b$ and $15 c$ are the first three terms of an AP, then
$$ \begin{aligned} & 2(7 b)=3 a+15 c \\ & \begin{array}{lcr} \Rightarrow & 14 a r=3 a+15 a r^{2} & {\left[\text { as } b=a r, c=a r^{2}\right]} \\ \Rightarrow & 14 r=3+15 r^{2} & {[\text { as } a \neq 0]} \end{array} \\ & \Rightarrow \quad 15 r^{2}-14 r+3=0 \\ & \Rightarrow \quad 15 r^{2}-5 r-9 r+3=0 \\ & \Rightarrow 5 r(3 r-1)-3(3 r-1)=0 \\ & \Rightarrow \quad(3 r-1)(5 r-3)=0 \\ & \Rightarrow \quad r=\frac{1}{3} \text { or } \frac{3}{5} \\ & \text { as, } r \in 0, \frac{1}{2} \text {, so } r=\frac{1}{3} \end{aligned} $$
Now, the common difference of $AP=7 b-3 a$
$$ =7 a r-3 a=a \frac{7}{3}-3=-\frac{2 a}{3} $$
So, $4^{\text {th }}$ term of $AP=3 a+3 \frac{-2 a}{3}=a$
