Question

If the ratio of the sum of first $n$ terms of two $A.Ps$ is $(7n+1):(4n+27)$, then find the ratio of their ${10}^{th}term$

Answer 1

As we know that, sum of $n$ terms in an A.P. is given as-

$S_n=\frac{n}{2}(2a+(n-1)d]$

Therefore,

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

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

For ratio of ${10}^{th}$ term,

$\frac{n-1}{2}=9$

$\Rightarrow n=18+1=19$

Substituting the value of $n$, in equation $\left(1\right)$, we have

$\frac{a_1+9d_1}{a_2+9d_2}=\frac{7\times 19+1}{4\times 19+27}=\frac{134}{103}$