The sums of $n$ terms of two arithmetic series are in the ratio of $7 n + 1 : 4 n + 27$ ; find the ratio of their $11^{t h}$ terms.

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Solution

Let the first term and common difference of the two series be $a_{1}, d_{1}$ and $a_{2}, d_{2}$ respectively.
We have $\frac{2 a_{1} + (n - 1) d_{1}}{2 a_{2} + (n - 1) d_{2}} = \frac{7 n + 1}{4 n + 27}$ .
Now we have to find the value of $\frac{a_{1} + 10 d_{1}}{a_{2} + 10 d_{2}}$ ; hence, by putting $n = 21$ , we obtain
$\frac{2 a_{1} + 20 d_{1}}{2 a_{2} + 20 d_{2}} = \frac{148}{111} = \frac{4}{3}$ ;
Thus, the required ratio is $4 : 3$ .

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