The sums of n terms of two arithmetic progressions are on the ratio 5n + 4 : 9n + 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 common differences of two A.P's respectively.

$∴ S_{n} = \frac{n}{2} [2 a_{1} + (n - 1) d_{1}]$

$S_{n} = \frac{n}{2} [2 a_{2} + (n - 1) d_{2}]$

$\frac{S_{n}}{S_{n}} = \frac{\frac{n}{2} [2 a_{1} + (n - 1) d_{1}]}{\frac{n}{2} [2 a_{2} + (n - 1) d_{2}]}$

= $\frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}}$

But $\frac{S_{n}}{S_{n}^{'}} = \frac{5 n + 4}{9 n - 6}$

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

= $\frac{a_{1} + (\frac{n - 1}{2}) d_{1}}{a_{2} + (\frac{n - 1}{2}) d_{2}}$

Now to get $18^{t h}$ terms $\frac{n - 1}{2} = 17$

$∴$ n = 35

Putting n = 35

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

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

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

Thus the ratio of $18^{t h}$ terms of two A.P. is 179 : 321.

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Similar questions

n

5 n + 4 : 9 n + 6

18^{t h}

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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