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

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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.

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

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

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

Substituting $n = 35$ in (1), we get
$\frac{{2 a}_{1} + 34 d_{1}}{{2 a}_{2} + 34 d_{2}} = \frac{5 (35) + 4}{9 (35) + 6}$

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

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

From eq (2) and eqn (3), we get
$\frac{18th term of first A.P.}{18th term of second A.P.} = \frac{179}{321}$
Ratio of $18^{t h}$ terms of two A.P.'s is $179 : 321$

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The sums of ' $n$ ' terms of two arithmetic progression are in the ration $5 n + 4 : 9 n + 6$ , find the ration of their $18^{t h}$ terms.

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