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Question
The sums of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their 18th terms.
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Solution
\[\text { Let there be two A . P . s } . \]
\[\text { Let their first terms be } a_1 \text { and }a_2 \text { and their common differences be } d_1 \text { and } d_2 . \]
\[\text { Given }: \]
\[ \frac{5n + 4}{9n + 6} = \frac{\text { Sum of n terms in the first A . P } .}{\text { Sum of n terms in the second A . P } .}\]
\[ \Rightarrow \frac{5n + 4}{9n + 6} = \frac{2 a_1 + [(n - 1) d_1 ]}{2 a_2 + [(n - 1) d_2 ]}\]
\[\text { Putting n } = 2 \times 18 - 1 = 35 \text { in the above equation, we get }: \]
\[ \frac{5 \times 35 + 4}{9 \times 35 + 6} = \frac{2 a_1 + 34 d_1}{2 a_2 + 34 d_2}\]
\[ \Rightarrow \frac{179}{321} = \frac{a_1 + 17 d_1}{a_1 + 17 d_1}\]
\[ \Rightarrow \frac{179}{321} = \frac{\text { 18th term of the first A . P } .}{\text { 18th term of the second A . P } .}\]
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