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 $A.P$.
$\therefore 2a_2=a_1+a_3...\left(1\right)$
Since $a_2,a_3,a_4$ are in $G.P.$
$\therefore a_3^2=a_2{a}_4...\left(2\right)$
Since $a_3,a_4,a_5$ are in $H.P.$
$\therefore a_4=\frac{2a_3{a}_5}{a_3+a_5}...\left(3\right)$
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 $G.P.$