Question

The sum of two numbers is $\frac{13}{6}$. An even number of arithmetic means are being inserted between them and their sum exceeds their number by $1$. Find the number of means inserted.

• A. $8$
• B. $10$
• C. $12$
• D. $14$

Answer 1

Answer: (C)

Let $a$ and $b$ be two numbers such that
$a+b=\frac{13}{6}\dots (i)$
Let $A_1,A_2,...,A_{2n}$ be $2n$ arithmetic means between $a$ and $b$.
then, $A_1+A_2+...+A_{2n}=2n\left(\frac{a+b}{2}\right)$
$\Rightarrow A_1+A_2+...+A_{2n}=n\left(a+b\right)=\frac{13}{6}n$ [Using(i)]
It is given that $A_1+A_2+...+A_{2n}=2n+1$
$\therefore \frac{13}{6}n=2n+1$
$\Rightarrow n=6$
So, number of means $=2n=12$