Question

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

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.

Given, $\frac{\text{Sum of n terms of first A.P.}}{\text{Sum of n terms of second A.P.}}=\frac{5n+4}{9n+6}$

$\Rightarrow \frac{\frac{n}{2}[{2a}_1+(n-1\left)d_1\right]}{\frac{n}{2}[{2a}_2+(n-1\left)d_2\right]}=\frac{5n+4}{9n+6}$

$\Rightarrow \frac{{2a}_1+(n-1)d_1}{{2a}_2+(n-1)d_2}=\frac{5n+4}{9n+6}$ ....(1)

Substituting $n=35$ in (1), we get

$\frac{{2a}_1+34d_1}{{2a}_2+34d_2}=\frac{5\left(35\right)+4}{9\left(35\right)+6}$

$\Rightarrow \frac{a_1+17d_1}{a_2+17d_2}=\frac{179}{321}$ ...(2)

Now, $\frac{\text{18th term of first A.P.}}{\text{18th term of second A.P.}}=\frac{a_1+17d_1}{a_2+17d_2}$ ....(3)

From eq (2) and eqn (3), we get

$\frac{\text{18th term of first A.P.}}{\text{18th term of second A.P.}}=\frac{179}{321}$

Ratio of ${18}^{th}$ terms of two A.P.'s is $179:321$