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Q.

The sum of n terms of two arithmetic progressions are in the ratio $(7 n + 1) : (4 n + 17)$ . then the ratio of their nth terms.

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a

$\frac{3 n + 2}{n - 1}$

b

$\frac{14 n - 6}{8 n + 23}$

c

non of these

d

$\frac{n - 1}{n}$

answer is A.

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Detailed Solution

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We have

$\begin{array}{l} \frac{s_{n}}{s_{n^{'}}} = \frac{7 n + 1}{4 n + 27} \\ \frac{s_{n}}{s_{n^{'}}} = \frac{7 n^{2} + n}{4 n^{2} + 27 n} \\ \frac{t_{n}}{t_{n^{'}}} = \frac{s_{n} - s_{n - 1}}{s_{n^{'}} - s_{n - 1^{'}}} \\ = \frac{(7 n^{2} + n) - (7 (n - 1)^{2} + (n - 1))}{(4 n^{2} + 27 n) - (4 (n - 1)^{2} + 27 (n - 1))} \\ = \frac{14 n - 6}{8 n + 23} \end{array}$

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The sum of n terms of two arithmetic progressions are in the ratio (7n+1):(4n+17). then the ratio of their nth terms.