Question

The sums of n terms of two arithmetic progressions are in the ratio $5n+4:9n+6$. Find the ratio of their ${18}^{th}$ terms.

Answer 1

For first AP

$n^{th}term=a_n=a+(n-1)d$

Sum of $n$ terms $=s_n=\frac{n}{2}(2a+(n-1\left)d\right)$

For second AP

$n^{th}term=A_n=A+(n-1)D$

Sum of $n$ terms $=S_n=\frac{n}{2}(2A+(n-1\left)D\right)$

We need to find $\frac{a_{18}}{A_{18}}=\frac{a+17d}{A+17D}$

It is Given that ,

$\frac{s_n}{S_n}=\frac{5n+4}{9n+6}\Rightarrow \frac{\frac{n}{2}(2a+(n-1\left)d\right)}{\frac{n}{2}(2A+(n-1\left)D\right)}=\frac{5n+4}{9n+6}\Rightarrow \frac{(2a+(n-1\left)d\right)}{(2A+(n-1\left)D\right)}=\frac{5n+4}{9n+6}$

$\Rightarrow \frac{2(a+\frac{n-1}{2}d)}{2(A+\frac{n-1}{2}D)}=\frac{5n+4}{9n+6}\Rightarrow \frac{(a+\frac{n-1}{2}d)}{(A+\frac{n-1}{2}D)}=\frac{5n+4}{9n+6}............\left(1\right)$

To Find $\frac{a+17d}{A+17D}$

$\frac{n-1}{2}=17\Rightarrow n=35$

Now in (1)

$\frac{a+\frac{35-1}{2}d}{A+\frac{35-1}{2}D}=\frac{5\times 35+4}{9\times 35+6}\Rightarrow \frac{a+17d}{A+17D}=\frac{179}{321}$

Therefore, the answer is $\frac{a_{18}}{A_{18}}=\frac{179}{321}$