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

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

Solution:

Let there be two A.P.s.

Let their first terms be $a_{1}$ and $a_{2}$ and their common differences be $d_{1}$ and $d_{2}$.

Given :

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

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

Putting $n=2 \times 18-1=35$ in the above equation, we get:

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

$\Rightarrow \frac{179}{321}=\frac{a_{1}+17 d_{1}}{a_{1}+17 d_{1}}$

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

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