The ratio of $n^{t h}$ term of two AP's is $(14 n - 6) : (8 n + 23)$ , then the ratio of their sum of first $m$ terms is

\frac{4 m + 4}{7 m + 24}

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\frac{7 m - 1}{4 m + 24}

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\frac{7 m + 1}{4 m + 27}

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\frac{28 m - 20}{16 m + 15}

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Solution

The correct option is C $\frac{7 m + 1}{4 m + 27}$
$\frac{T_{n}}{T_{n}^{'}} = \frac{a_{1} + (n - 1) d_{1}}{a_{1}^{'} + (n - 1) d_{1}^{'}} = \frac{14 n - 6}{8 n + 23}$
comparing coff.of n & constant terms
$d_{1} = 14, a_{1} - d_{1} = - 6 \Rightarrow a_{1} = 8$
$d_{1}^{'} = 8, a_{1}^{'} - d_{1}^{'} = 23 \Rightarrow a_{1}^{'} = 31$
Ratio of sum of $1^{s t}$ m terms $= \frac{m / 2 (2 a_{1} + (m - 1) d_{1})}{m / 2 (2 a_{1}^{'} + (m - 1) d_{1}^{'})}$
$= \frac{2 a_{1} + (m - 1) d_{1}}{2 a_{1}^{'} + (m - 1) d_{1}^{'}} = \frac{16 + (m - 1) 14}{62 + (m - 1) 8}$
$= \frac{2 + 14 m}{54 + 8 m} = \frac{1 + 7 m}{4 m + 27} - (C)$

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