If $a_{1}$ , $a_{2}$ , $a_{3}$ are in AP $a_{2}$ , $a_{3}$ , $a_{4}$ are in GP and $a_{3}$ , $a_{4}$ , $a_{5}$ are in HP then $a_{1}$ , $a_{3}$ , $a_{5}$ are in

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Solution

The correct option is A GP
Given $a_{1}, a_{2},, a_{3}$ are in A.P
Let's assume that $a_{1} = A - d a_{2} = A a_{3} = A + d$
similarly $a_{2}, a_{3}, a_{4}$ are in G.P
$s o a_{2} = A a_{3} = A r a_{4} = A r^{2}$
$\Rightarrow A + d = A r \Rightarrow d = A r - A$
and finally $a_{3}, a_{4}, a_{5}$ are in H.P
$\Rightarrow \frac{1}{a_{3}}, \frac{1}{a_{4}}, \frac{1}{a_{5}}$ are in A.P
hence $\Rightarrow \frac{1}{a_{5}} = \frac{1}{a_{4}} + (\frac{1}{a_{4}} - \frac{1}{a_{3}}) \Rightarrow \frac{1}{a_{5}} = \frac{2}{A r^{2}} - \frac{1}{A r} = \frac{2 - r}{A r^{2}} \Rightarrow a_{5} = \frac{A r^{2}}{2 - r}$
and $a_{1} = A - (A r - A) = 2 A - A r = A (2 - r) a_{3} = A r$
clearly we can see that $a_{1} a_{5} = {a_{3}}^{2}$
Thus $a_{1}, a_{3}, a_{5}$ are in G.P

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

Let a $_{1}$ , a $_{2}$ , a $_{3}$ , a $_{4}$ and a $_{5}$ be such that a $_{1}$ , a $_{2}$ and a $_{3}$ are in A.P., a $_{2}$ , a $_{3}$ and a $_{4}$ are in G.P., and a $_{3}$ , a $_{4}$ and a $_{5}$ are in H.P. Then log $_{e}$ a $_{1}$ , log $_{e}$ $a_{3}$ and log $_{e}$ a $_{5}$ are in