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Question
If the sums of n terms of two arithmetic progressions are in the ratio 2n + 5 : 3n + 4, then write the ratio of their m th terms.
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Solution
Given:
\[\frac{S_n}{{S_n}^1} = \frac{2n + 5}{3n + 4}\]
\[ \Rightarrow \frac{\frac{n}{2}\left\{ 2a + \left( n - 1 \right)d \right\}}{\frac{n}{2}\left\{ 2b + \left( n - 1 \right) d^1 \right\}} = \frac{2n + 5}{3n + 4}\]
\[ \Rightarrow \frac{2a + \left( n - 1 \right)d}{2b + \left( n - 1 \right) d^1} = \frac{2n + 5}{3n + 4} . . . \left( 1 \right)\]
Ratio of their m terms =\[\frac{a_m}{b_m}\]
To find the ratio of the mth terms, replace n by 2m \[-\] 1 in equation (1).
\[\Rightarrow \frac{2a + \left( 2m - 2 \right)d}{2b + \left( 2m - 2 \right) d^1} = \frac{2\left( 2m - 1 \right) + 5}{3\left( 2m - 1 \right) + 4}\]
\[ \Rightarrow \frac{a + \left( m - 1 \right)d}{b + \left( m - 1 \right) d^1} = \frac{4m - 2 + 3}{6m - 3 + 4}\]
\[ \Rightarrow \frac{a_m}{b_m} = \frac{4m + 1}{6m + 1}\]
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