Question

The sums of $n$ terms of two arithmetic progress are in the ratio $5n+4:9n+6$. Find the ratio of their ${18}^{th}$ terms.

Answer 1

Let $a_1,d_1$ be the first term and common difference of the first A.P. respectively
and $a_2,d_2$ be the first term and common difference of the second A.P. respectively

According to given condition, we have
$\frac{{\text{Sum of n terms A.P}}_1}{{\text{Sum of n terms A.P}}_2}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{\frac{n}{2}[2a_1+(n-1\left)d_1\right]}{\frac{n}{2}[2a_2+(n-1\left)d_2\right]}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2]}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{2[a_1+\frac{(n-1)}{2}{d}_1]}{2[a_2+\frac{(n-1)}{2}{d}_2]}=\frac{5n+4}{9n+6}$
$\Rightarrow \frac{a_1+\frac{(n-1)}{2}{d}_1}{a_2+\frac{(n-1)}{2}{d}_2}=\frac{5n+4}{9n+6}\cdots \left(i\right)$

Finding the ratio of their ${18}^{th}$ terms.
Substitute the coefficient of common difference equal to $17$.
$\therefore \frac{n-1}{2}=17$
$\Rightarrow n-1=34$
$\therefore n=35$ substitute in equation $\left(i\right)$
$\therefore \frac{a_1+17d_1}{a_2+17d_2}=\frac{5\times 35+4}{9\times 35+6}$

$\Rightarrow \frac{{18}^{th}{\text{term of A.P}}_1}{{18}^{th}{\text{term of A.P}}_2}=\frac{179}{321}$

Final answer: Hence, the required ratio of ${18}^{th}$ term is $179:321$.