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Question
The sums of first n terms of two A.P.'s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5th terms.
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Solution
\[\text { Let the first term, the common difference and the sum of the first n terms of the first A . P . be} a_1 , d_1 and S_1 , respectively, and those of the second A . P . be a_2 , d_2 and S_2 , respectively . \]
\[\text { Then, we have }, \]
\[ S_1 = \frac{n}{2}\left[ 2 a_1 + \left( n - 1 \right) d_1 \right] \]
\[\text { And, } S_2 = \frac{n}{2}\left[ 2 a_2 + \left( n - 1 \right) d_2 \right]\]
\[\text { Given: } \]
\[ \frac{S_1}{S_2} = \frac{\frac{n}{2}\left[ 2 a_1 + \left( n - 1 \right) d_1 \right]}{\frac{n}{2}\left[ 2 a_2 + \left( n - 1 \right) d_2 \right]} = \frac{7n + 2}{n + 4}\]
\[ \Rightarrow \frac{S_1}{S_2} = \frac{\left[ 2 a_1 + \left( n - 1 \right) d_1 \right]}{\left[ 2 a_2 + \left( n - 1 \right) d_2 \right]} = \frac{7n + 2}{n + 4}\]
\[\text { To find the ratio of the 5th terms of the two A . P . s, we replace n by } (2 \times 5 - 1 = 9)\text { in the above equation }: \]
\[ \Rightarrow \frac{\left[ 2 a_1 + \left( 9 - 1 \right) d_1 \right]}{\left[ 2 a_2 + \left( 9 - 1 \right) d_2 \right]} = \frac{7 \times 9 + 2}{9 + 4}\]
\[ \Rightarrow \frac{\left[ 2 a_1 + \left( 8 \right) d_1 \right]}{\left[ 2 a_2 + \left( 8 \right) d_2 \right]} = \frac{7 \times 9 + 2}{9 + 4} = \frac{65}{13} \]
\[ \Rightarrow \frac{\left[ a_1 + 4 d_1 \right]}{\left[ a_2 + 4 d_2 \right]} = \frac{5}{1} = 5: 1\]
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