If the ratio of the sum of n terms of two AP's be (7n + 1):(4n + 27), then the ratio of their 11th terms will be

Let S_n, and S' be the sums of n terms of two AP's and T₁₁ and T₁₁, be the respective 11th term, then

$\frac{S_{n}}{S_{n}^{'}} = \frac{\frac{n}{2} [2 a + (n - 1) d]}{\frac{n}{2} [2 a^{'} + (n - 1) d^{'}]} = \frac{7 n + 1}{4 n + 27}$ (given)

$\Rightarrow$ $\frac{a + \frac{(n - 1)}{2} d}{a^{'} + \frac{(n - 1)}{2} d^{'}} = \frac{7 n + 1}{4 n + 27}$

Now, put n=21, we get

$\frac{a + 10 d}{a^{'} + 10 d^{'}} = \frac{T_{11}}{T_{11}^{'}} = \frac{148}{111} = \frac{4}{3}$

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