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
$a+b= \frac{13}{6}\quad \ldots (i)$
Let $A_{1}, A_{2}, ...,A_{2n}$ be $2n$ arithmetic means between $a$ and $b$.
then, $A_{1}+A_{2} +...+A_{2n}= 2n\left(\frac{a+b}{2}\right)$
$\Rightarrow A_{1} +A_{2} +...+A_{2n} = n\left(a+b\right)= \frac{13}{6}n \quad$ [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 $