Question

If $\frac{SumofntermsoffirstA.P.}{SumofntermsofsecondA.P.}=\frac{5n+4}{9n+6}$. Find the ratio of their ${18}^{th}$ term.

Answer 1

Let $a_1,a_2,$ and $d_1,d_2$ be the first terms and the common difference of the first and second arithmetic progression respectively.
According to the given condition,
$\frac{SumofntermsoffirstA.P.}{SumofntermsofsecondA.P.}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}=\frac{5n+4}{9n+6}\Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{5n+4}{9n+6}.....\left(1\right)$
Substituting n =35 in (1), we obtain
$\frac{2a_1+34d_1}{2a_2+34d_2}=\frac{5\left(35\right)+4}{9\left(35\right)+6}\Rightarrow \frac{a_1+17d_1}{a_2+17d_2}=\frac{179}{321}....\left(2\right)$
$\frac{{18}^{th}termoffirstA.P.}{{18}^{th}termofsecondA.P}=\frac{a_1+17d_1}{a_2+17d_2}....\left(3\right)$
From (2) and (3), we obtain
$\frac{{18}^{th}termoffirstA.P}{{18}^{th}termofsecondA.P.}=\frac{179}{321}$
Thus, the ratio of ${18}^{th}$ term of both the A.P.s is 179:321