Question

If the ratio of the ${11}^{th}$ term of an A.P. to its ${18}^{th}$ term is $2:3$, find the ratio of the sum of first five terms to the sum of its first 10 terms

Answer 1

Given ${11}^{th}$ term of an A.P$=T_{11}=a+10d$

and ${18}^{th}$ term of an A.P$=T_{18}=a+17d$

Ratio$=\frac{a+10d}{a+17d}=\frac{2}{3}$

$\Rightarrow 3(a+10d)=2(a+17d)$

$\Rightarrow 3a+30d=2a+34d$

$\Rightarrow 3a-2a=34d-30d=4d$

$\therefore a=4d$

We have Sum of $n$ terms of an A.P$=S_n=\frac{n}{2}(2a+(n-1)d)$

Again $\frac{S_5}{S_{10}}=\frac{\frac{5}{2}(2\times 4d+(5-1)d)}{\frac{10}{2}(2\times 4d+(10-1)d)}$

$=\frac{5(8d+4d)}{10(8d+9d)}$

$=\frac{5\times 12d}{10\times 17d}=\frac{6}{17}$

Thus, the ratio is $6:17$