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If 3 + 5 + 7 + . . . + Upto N Terms 5 + 8 + 11 + . . . . Upto 10 Terms 7, Then Find the Value of N.

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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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Chapter 19: Arithmetic Progression - Exercise 19.8 [Page 51]

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R.D. Sharma Mathematics [English] Class 11
Chapter 19 Arithmetic Progression
Exercise 19.8 | Q 9 | Page 51

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