Question

Let there are $2$ numbers whose sum is $\frac{13}{6}$. The sum of the $A{M}^{\prime }s$ between them which are even in number exceeds the number of means by $2$. Then the number of means inserted is-

• A. $10$
• B. $12$
• C. $8$
• D. $24$

Answer 1

The correct option is D $24$

Let a and b be two mumbers and

A.M.'s are inserted betwem a and b.

Then,

$\frac{2n}{2}(a+b)=2n+2\left(given\right)$

$\Rightarrow n\times \frac{13}{6}=2(n+1)(a+b=\frac{13}{6})$

$\Rightarrow 13n=12n+12$

$\Rightarrow n=12$

$\therefore$ No of means inserted $=2n$

$=2\times 12$

$=24$

$\therefore$ option $D$ is correct