Question

The ratio between the sum of n terms of two A.P.'s is $3n+8:7n+15$. Find the ratio between their $12$th terms.

Answer 1

$\frac{S_n}{S_n^{\prime }}=\frac{\left(n/2\right)[2a+(n-1\left)d\right]}{\left(n/2\right)[2d+(n-1\left)d^{\prime }\right]}=\frac{3n+8}{7n+15}$
or $\frac{a+\left[\right(n-1\left)/2\right]d}{d+\left[\right(n-1\left)/2\right]d^{\prime }}=\frac{3n+8}{7n+15}$ .$\left(1\right)$
We have to find $\frac{T_{12}}{T_{12}^{\prime }}=\frac{a+11d}{a^{\prime }+11d^{\prime }}$
Choosing $(n-1)/2=11$ or $n=23$ in $\left(1\right)$,
We get
$\frac{T_{12}}{T_{12}^{\prime }}=\frac{a+11d}{a^{\prime }+11d^{\prime }}=\frac{3\left(23\right)+8}{7\left(23\right)+15}=\frac{77}{176}=\frac{7}{16}$.