Question

Between two numbers whose sum is $2\frac{1}{6}$ an even number of arithmetic means are inserted; the sum of these means exceeds their number by unity : how many means are there?

Answer 1

The sum of two numbers is $\frac{13}{6}$

Let the numbers be $A$ and $\frac{13}{6}-A$

Let $2n$ means be inserted

$\therefore$ Sum of $2n+2$ terms $=\frac{(No.ofterms)}{2}(first+lastterm)$

$=\frac{(2n+2)}{2}\times [A+\frac{13}{6}-A]=\frac{13}{6}(n+1)$

Sum of $2n$ means$=$ (sum of $(2n+2)$terms)$-$(Sum of 1st and last term)

$=\frac{13}{6}(n+1)-\frac{13}{6}=\frac{13n}{6}$

$\therefore$ This is more than their number $2n$ by $1=(2n+1)$

$\therefore \frac{13n}{6}=2n+1$

$⟹13n=12n+6$

$⟹n=6$