Question

The sum of $n$ terms of two A.Ps. are in ratio $\frac{7n+1}{4n+27}.$ Find the ratio of their ${11}^{th}$ terms.

Answer 1

Let $a_1$ and $a_2$ be the first terms and $d_1$ and $d_2$ be the common differences of two A.P.s respectively, then

By using $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{7n+1}{4n+27}$

$\Rightarrow$ $\frac{a_1+\left(\frac{n-1}{2}\right)d_1}{a_2+\left(\frac{n-1}{2}\right)d_2}=\frac{7n+1}{4n+27}$

For ratio of ${11}^{th}$ terms

$\frac{n-1}{2}=10\Rightarrow n=21$

so ratio of ${11}^{th}$ terms is$=\frac{7\left(21\right)+1}{4\left(21\right)+27}=\frac{148}{111}=\frac{4}{3}$