Sequences And Series Ques 3
- If $m$ is the AM of two distinct real numbers $l$ and $n(l, n>1)$ and $G_1, G_2$ and $G_3$ are three geometric means between $l$ and $n$, then $G_1^4+2 G_2^4+G_3^4$ equals
(2015)
(a) $4 $ $l^2 m n$
(b) $4 \mathrm{lm}^2 n$
(c) $l m n^2$
(d) $l^2 m^2 n^2$
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Answer:
Correct Answer: 3.(b)
Solution: (b) Given, $m$ is the AM of $l$ and $n$.
$ \therefore \quad l+n=2 m $
and $G_1, G_2, G_3$ are geometric means between $l$ and $n$.
$l, G_1, G_2, G_3, n$ are in GP.
Let $r$ be the common ratio of this GP.
$ \therefore G_1=l r, G_2=l r^2, G_3=l r^3, n-l r^4 \Rightarrow r=\left(\frac{n}{l}\right)^{\frac{1}{4}} $
Now, $G_1^4+2 G_2^4+G_3^4=(l r)^4+2\left(l r^2\right)^4+\left(l r^3\right)^4$
$ \begin{aligned} & =l^4 \times r^4\left(1+2 r^4+r^6\right)=l^4 \times r^4\left(r^4+1\right)^2 \\ & =l^4 \times \frac{n}{l}\left(\frac{n+l}{l}\right)^2=l n \times 4 m^2=4 l m^2 n \end{aligned} $
BETA
