Question

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, $a_1,a_3$ and $a_5$ are in

• A. G.P
• B. A.P
• C. H.P
• D. none of these.

Answer 1

Answer: (A)

Since $a_{1}, a_{2}, a_{3}$ are in $A.P$.
$\therefore 2a_{2} = a_{1}+a_{3} \quad...\left(1\right)$
Since $a_{2}, a_{3}, a_{4} $ are in $G.P. $
$\therefore a_{3}^{2} = a_{2}a_{4} \quad...\left(2\right) $
Since $a_{3}, a_{4}, a_{5}$ are in $H.P.$
$\therefore a_{4} = \frac{2a_{3}a_{5}}{a_{3}+a_{5}} \quad...\left(3\right)$
Putting $a_{2} = \frac{a_{1}+a_{3}}{2}$ and $a_{4}= \frac{2a_{3}a_{5}}{a_{3}+a_{5}}$ in $\left(2\right)$,
We get $a_{3}^{2} = \frac{a_{1}+a_{3}}{2}\cdot \frac{2a_{3}a_{5}}{a_{3}+a_{5}} $
$a_{3}^{2} + a_{3}a_{5} = a_{1}a_{5} +a_{3}a_{5} $
$ \Rightarrow a_{3}^{2} = a_{1}a_{5}$
$ \Rightarrow a_{1}, a_{3}, a_{5}$ are in $G.P.$