Question

' $n$ ' arithmetic means $A_{1}, A_{2} \ldots, A_{n}$ are inserted between $7$ and $49$ such that sum of these means is $364$ . If the $G.M$. of $A_{3}$ and $A_{6}$ is $k$, then $\left(\frac{k}{4}\right)$ is equal to

Answer 1

We have $A_{1}+A_{2}+\ldots+A_{n}=n\left(\frac{7+49}{2}\right)=364$
$\Rightarrow n=13$
Hence $49=7+14 d \Rightarrow d=3$
$\therefore A_{1}=7+3=10, A_{2}=13, A_{3}=16, \ldots, A_{6}=25 $
$\Rightarrow k=\sqrt{16.25}=20 .$