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(A) Let the sum of $n$ terms of two arithmetic progressions be $S_n$ and $S'_n$,and their $11^{th}$ terms be $T_{11}$ and $T'_{11}$ respectively.$\fra

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$4 : 3$

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(A) Let the sum of $n$ terms of two arithmetic progressions be $S_n$ and $S'_n$,and their $11^{th}$ terms be $T_{11}$ and $T'_{11}$ respectively.$\frac{S_n}{S'_n} = \frac{\frac{n}{2}[2a + (n - 1)d]}{\frac{n}{2}[2a' + (n - 1)d']} = \frac{7n + 1}{4n + 27}$$\frac{a + \frac{n - 1}{2}d}{a' + \frac{n - 1}{2}d'} = \frac{7n + 1}{4n + 27}$To find the ratio of the $11^{th}$ terms,we need $\frac{a + 10d}{a' + 10d'}$.Setting $\frac{n - 1}{2} = 10$,we get $n - 1 = 20$,so $n = 21$.Substituting $n = 21$ in the ratio:$\frac{T_{11}}{T'_{11}} = \frac{7(21) + 1}{4(21) + 27} = \frac{147 + 1}{84 + 27} = \frac{148}{111} = \frac{4}{3}$.

If the ratio of the sum of $n$ terms of two arithmetic progressions is $(7n + 1) : (4n + 27)$,then what is the ratio of their $11^{th}$ terms?

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