The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.
Question:

The sums of $n$ terms of two arithmetic progressions are in the ratio $5 n+4: 9 n+6$. Find the ratio of their $18^{8 \mathrm{~h}}$ terms.

Solution:

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,

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

$\Rightarrow \frac{\frac{n}{2}\left[2 a_{1}+(n-1) d_{1}\right]}{\frac{n}{2}\left[2 a_{2}+(n-1) d_{2}\right]}=\frac{5 n+4}{9 n+6}$

$\Rightarrow \frac{2 a_{1}+(n-1) d_{1}}{2 a_{2}+(n-1) d_{2}}=\frac{5 n+4}{9 n+6}$ (1)

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

$\frac{2 a_{1}+34 d_{1}}{2 a_{2}+34 d_{2}}=\frac{5(35)+4}{9(35)+6}$

$\Rightarrow \frac{a_{1}+17 d_{1}}{a_{2}+17 d_{2}}=\frac{179}{321}$ …(2)

$\frac{18^{\text {th }} \text { term of first A.P. }}{18^{\text {th }} \text { term of second A.P }}=\frac{a_{1}+17 d_{1}}{a_{2}+17 d_{2}}$ (3)

From (2) and (3), we obtain

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

Thus, the ratio of $18^{\text {th }}$ term of both the A.P.S is $179: 321$.