Question

If the $A.M$., the $G.M$. and the $H.M$. of the first and the last terms of the series $100,101,102,.........,n$ are the terms of the series itself then the sum of digit in $n$ is $(100<n\le 500)$

• A. $4$
• B. $400$
• C. $100$
• D. None of these

Answer 1

The correct option is A $4$

The first and the last terms of the series are $100$ and $n$

A.M. $=\frac{100+n}{2}$, G.M. $=10\sqrt{n}$

And
H.M. $=\frac{200n}{100+n}$

For A.M. to be an integer $n$ must be even

For G.M. to be an integer $n$ must be perfect square

$\Rightarrow n=4k^2$

$\Rightarrow$H.M $=\frac{200k^2}{25+k^2}$

Since numerator is multiple of $25$, denominator must be a multiple of $25$ as H.M. is an integer.

Also $100<n<500$

$\Rightarrow 25<k^2\le 125$

$\to k^2=50,75,100$ and $125$

Satisfying these values, we get

$k^2=100$ as the only solution.

$\Rightarrow n=400$

Sum =4