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Question
If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3), then find the ratio of their 12th terms.
Solution
Let the first terms of the two A.P.'s be a and a'; and their common difference be d and d'.
Now,
\[\frac{S_n}{S_n '} = \frac{\left( 3n + 2 \right)}{\left( 2n + 3 \right)}\]
\[ \Rightarrow \frac{\frac{n}{2}\left[ 2a + \left( n - 1 \right)d \right]}{\frac{n}{2}\left[ 2a' + \left( n - 1 \right)d' \right]} = \frac{\left( 3n + 2 \right)}{\left( 2n + 3 \right)}\]
\[ \Rightarrow \frac{\left[ 2a + \left( n - 1 \right)d \right]}{\left[ 2a' + \left( n - 1 \right)d' \right]} = \frac{\left( 3n + 2 \right)}{\left( 2n + 3 \right)}\]
\[\text { Let }n = 23\]
\[ \Rightarrow \frac{2a + \left( 23 - 1 \right)d}{2a' + \left( 23 - 1 \right)d'} = \frac{3 \times 23 + 2}{2 \times 23 + 3}\]
\[ \Rightarrow \frac{2a + 22d}{2a' + 22d'} = \frac{69 + 2}{46 + 3}\]
\[ \Rightarrow \frac{2\left( a + 11d \right)}{2\left( a' + 11d' \right)} = \frac{71}{49}\]
\[ \therefore \frac{a_{12}}{a_{12'} } = \frac{71}{49}\]
So, the ratio of their 12th terms is 71 : 49.
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