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Question
If \[\frac{3 + 5 + 7 + . . . + \text { upto n terms }}{5 + 8 + 11 + . . . . \text { upto 10 terms }}\] 7, then find the value of n.
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Solution
\[\frac{3 + 5 + 7 + . . . + \text { upto n terms }}{5 + 8 + 11 + . . . . \text { upto 10 terms }} = 7\]
\[ \Rightarrow \frac{\frac{n}{2}\left\{ 2 \times 3 + \left( n - 1 \right)2 \right\}}{\frac{10}{2}\left\{ 2 \times 5 + \left( 10 - 1 \right)3 \right\}} = 7\]
\[ \Rightarrow \frac{n\left( 4 + 2n \right)}{370} = 7\]
\[ \Rightarrow n^2 + 2n - 1295 = 0\]
\[ \Rightarrow n^2 + 37n - 35n - 1295 = 0\]
\[ \Rightarrow \left( n + 37 \right)\left( n - 35 \right)\]
\[ \Rightarrow n = 35, n = - 37\]
\[\text { Rejecting the negative value, we get }: \]
\[ \Rightarrow n = 35\]
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