Question

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

Answer 1

Simplication of ratio of their $n^{th}$ terms.
Let $(A.P.{)}_1\to a_1$ and $d_1$ are the first term and common difference respectively
$(A.P.{)}_2\to a_2$ and $d_2$ are the first term and common difference respectively

According to given condition, we have
$\frac{{\text{Sum of}n\text{terms of A.P.}}_1}{{\text{Sum of}n\text{terms of A.P.}}_2}=\frac{(3n+8)}{(7n+15)}$
$\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}=\frac{3n+8}{7n+15}$
$\Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+8}{7n+15}$
$\Rightarrow \frac{2[a_1+\frac{n-1}{2}{d}_1]}{2[a_2+\frac{n-1}{2}{d}_2]}=\frac{3n+8}{7n+15}$
$\Rightarrow \frac{a_1+\frac{(n-1)}{2}{d}_1}{a_2+\frac{(n-1)}{2}{d}_2}=\frac{3n+8}{7n+15}\cdots \left(i\right)$

Finding the ratio of their ${12}^{th}$ terms.
Substitute the coefficient of common difference to $11$.
$\therefore \frac{n-1}{2}=11$
$\Rightarrow n-1=22$
$\Rightarrow n=23$ substitute this in equation $\left(i\right)$
$\therefore \frac{a_1+11d_1}{a_2+11d_2}=\frac{3\times 23+8}{7\times 23+15}$
$\frac{a_1+11d_1}{a_2+11d_2}=\frac{77}{176}$

$\therefore \frac{{12}^{th}\text{term of}A.P{.}_1}{{12}^{th}\text{term of}A.P{.}_2}=\frac{7}{16}$

Final answer:
Hence, the required ratio is $7:16$.