Question

The sum of n terms of two A.P. are in the ratio 5n+4:9n+6.
The ratio of their 18th terms is

• A. 47:84
• B. 179:321
• C. 89:159
• D. 2:3

Answer 1

The correct option is B 179:321

Let the sum of nth terms of two AP be denoted by $s_n\mathrm{and}{S}_n$. The first term and common difference of two AP's be denoted by a & d and A & D respectively
$\frac{s_n}{S_n}=\frac{5n+4}{9n+6}\Rightarrow \frac{\frac{n}{2}\left(2a+(n-1)d\right)}{\frac{n}{2}\left(2A+(n-1)D\right)}=\frac{5n+4}{9n+6}\Rightarrow \frac{\left(}{2a+(n-1)d}=\frac{5n+4}{9n+6}\mathrm{Dividing}\mathrm{Numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}2,\frac{a+\frac{(n-1)}{2}d}{A+\frac{(n-1)}{2}D}=\frac{5n+4}{9n+6}$
The ratio of 18th term of two A.P. is
$\frac{a_{18}}{A_{18}}=\frac{a+17d}{A+17D}\mathrm{Comparing}\mathrm{with}\frac{a+\frac{(n-1)}{2}d}{A+\frac{(n-1)}{2}D}=\frac{5n+4}{9n+6}...\left(1\right]\mathrm{we}\mathrm{get},\frac{(n-1)}{2}=17\Rightarrow n=35\mathrm{Putting}n=35\mathrm{in}\left(1\right],\mathrm{we}\mathrm{get}\frac{a_{18}}{A_{18}}=\frac{5\left(35\right)+4}{9\left(35\right)+6}=\frac{179}{321}$