Sequences and Series 3 Question 5
5. Let $a, b$ and $c$ be the 7th, 11th and 13 th terms respectively of a non-constant AP. If these are also the three consecutive terms of a GP, then $\frac{a}{c}$ is equal to
(a) 2
(b) $\frac{7}{13}$
(c) 4
(d) $\frac{1}{2}$
(2019 Main, 9 Jan II)
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Answer:
Correct Answer: 5. (b)
Solution:
- Let $A$ be the Ist term of $AP$ and $d$ be the common difference.
$$ \therefore \quad 7 \text { th term }=a=A+6 d $$
| Then, $\Rightarrow$ |
$a+b+c=x b$ $a+a r+a r^{2}=x a r$ |
|
|---|---|---|
 |
$1+r+r^{2}=x r$ | $\ldots$ (i) $[\because a \neq 0]$ |
| $x=\frac{1+r+r^{2}}{=1+r+\underline{1}}$ |
$[\because n$th term $=A+(n-1) d]$ 11 th term $=b=A+10 d$
13 th term $=c=A+12 d$
$\because a, b, c$ are also in GP
$\therefore b^{2}=a c$
$\Rightarrow(A+10 d)^{2}=(A+6 d)(A+12 d)$
$\Rightarrow A^{2}+20 A d+100 d^{2}=A^{2}+18 A d+72 d^{2}$
$\Rightarrow 2 A d+28 d^{2}=0$
$\Rightarrow 2 d(A+14 d)=0$
$\Rightarrow d=0$ or $A+14 d=0$
But $\quad d \neq 0$
$[\because$ the series is non constant AP]
$\Rightarrow \quad A=-14 d$
$\therefore \quad a=A+6 d=-14 d+6 d=-8 d$
and $\quad c=A+12 d=-14 d+12 d=-2 d$
$\Rightarrow \quad \frac{a}{c}=\frac{-8 d}{-2 d}=4$
