Sequences and Series 3 Question 14
14.
Find three numbers $a, b, c$ between 2 and 18 such that (i) their sum is 25 . (ii) the numbers $2, a, b$ are consecutive terms of an AP. (iii) the numbers $b, c, 18$ are consecutive terms of a GP.
(1983, 2M)
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Answer:
Correct Answer: 14. $(a=5)(b=8)(c=12)$
Solution:
- If $a, b, c \in(2,18)$, then
$ a+b+c=25 $
Since, $2, a, b$ are in AP.
$\Rightarrow \quad 2 a=b+2$
and $b, c, 18$ are in GP.
$\Rightarrow \quad c^{2}=18 b$
From Eqs. (i), (ii) and (iii),
$ \begin{aligned} & \frac{b+2}{2}+b+\sqrt{18 b}=25 \\ & \Rightarrow \quad 3 b+2+6 \sqrt{2} \sqrt{b}=50 \\ & \Rightarrow \quad 3 b+6 \sqrt{2} \sqrt{b}-48=0 \\ & \Rightarrow \quad b+2 \sqrt{2} \sqrt{b}-16=0 \\ & \Rightarrow \quad b+4 \sqrt{2} \sqrt{b}-2 \sqrt{2} \sqrt{b}-16=0 \\ & \Rightarrow \sqrt{b}(\sqrt{b}+4 \sqrt{2})-2 \sqrt{2}(\sqrt{b}+4 \sqrt{2})=0 \\ & \Rightarrow \quad(\sqrt{b}-2 \sqrt{2})(\sqrt{b}+4 \sqrt{2})=0 \\ & \Rightarrow \quad b=8, a=5 \\ & \text { and } \quad c=12 \end{aligned} $
