Question

If the sum of the first $3n$ terms is equal to the next $n$ term of an $A.P$ whose common difference is non zero then find the reiprocal of ratio of the sum of the first $2n$ terms to the next $2n$ terms ?

Answer 1

Given,

sum of first $3n$ terms=sum of the next $n$ terms

We know that, in an $A.P$

sum of the first $n$ terms, $S_n=\left(\frac{n}{2}\right)\{2a+(n-1\left)d\right\}$ and the $nth$ term= $a+(n-1)d$

So, $\left(\frac{3n}{2}\right)\{2a+(3n-1\left)d\right\}=\left(\frac{n}{2}\right)[2\{a+(3n-1-d\left)\right\}+(n-1\left)d\right]\Rightarrow 6a+9nd-3d=2a+6nd+nd-d\Rightarrow 4a+2nd-2d=0\Rightarrow 2a+nd-d=0⟶\left(1\right)$

The ratio of the sum of the first $2n$ terms to the next $2n$ terms,

$\frac{\left(\frac{2n}{2}\right)\{2a+(2n-1\left)d\right\}}{\left(\frac{2n}{2}\right)[2\left\{\right(a+2n+1-1\left)d\right\}+(2n-1\left)d\right]}=\frac{(2a+2nd-d)}{(2a+4nd+2nd-d)}=\frac{(2a+nd-d+nd)}{(2a+nd-d+5nd)}=\frac{(0+nd)}{(0+5nd)}[\because 2a+nd-d=0]=\frac{nd}{5nd}=\frac{1}{5}$

Hence, the reciprocal of ratio of the sum of first $2n$ terms to the next $2n$ terms is $5:1.$