If $${{{1^3} + {2^3} + {3^3}\, + \,...\,up\,to\,n\,terms} \over {1\,.\,3 + 2\,.\,5 + 3\,.\,7\, + \,...\,up\,to\,n\,terms}} = {9 \over 5}$$, then the value of $n$ is
Answer (integer)
5
Solution
Given $\frac{1^{3}+2^{3}+3^{3}+\ldots \text { up to } n \text { terms }}{1.3+2.5+3.7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$
<br/><br/>
Now
<br/><br/>
Let $S=1.3+2.5+3.7+\ldots$
<br/><br/>
$$
\begin{aligned}
& T_{n}=n \cdot(2 n+1) \\\\
& \therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\\\
& \Rightarrow \frac{\left(\frac{n(n+1)}{2}\right)^{2}}{n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]}=\frac{9}{5} \\\\
& \Rightarrow 5 n^{2}-19 n-30=0 \\\\
& \Rightarrow(5 n+6)(n-5)=0 \\\\
& \therefore n=5
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
This question is part of PrepWiser's free JEE Main question bank. 209 more solved questions on Sequences and Series are available — start with the harder ones if your accuracy is >70%.