The sum of $n$ terms of two arithmetic progressions are in the ratio $\left(3n+8\right):\left(7n+15\right)\dot{}$ Find the ratio of their 12th terms.

Solution: Let $a$ is the first term and $d_1$ is the common difference of the first AP.
Let $b$ is the first term and $d_2$ is the common difference of the second AP.
Then,
$\frac{\frac{n}{2}\left(2a+\left(n-1\right)d_1\right)}{\frac{n}{2}\left(2b+\left(n-1\right)d_2\right)}=\frac{3n+8}{7n+15}$
$\Rightarrow \frac{2a+\left(n-1\right)d_1}{2b+\left(n-1\right)d_2}=\frac{k\left(3n+8\right)}{k\left(7n+15\right)}$
Comparing numerators for both sides,
$\Rightarrow 2a-d_1=8k,d_1=3k$
$\Rightarrow a=\frac{11}{2}k$
Similarly, comparing denominators of both sides,
$\Rightarrow 2b-d_2=15k,d_2=7k$
$\Rightarrow b=11k$
So, ratio of their ${12}^{th}$ terms,
$\frac{a_{12}}{b_{12}}=\frac{a+11d_1}{b+11d_2}$
$=\frac{\frac{11}{2}k+33k}{11k+77k}=\frac{77k}{176k}=\frac{7}{16}$
So, required ratio is $\frac{7}{16}$.