Question

If $\frac{3+5+7+....n}{5+8+11+....10}=7$, then the value of $n$ is

• A. $35$
• B. $36$
• C. $37$
• D. $40$

Answer 1

The correct option is A

$35$

Explanation for the correct option:

Step 1. First take $\left(3+5+7+....n\right)$

It is an A.P with $a=3$ and $d=2$

As we know,

Sum of $n$ terms of an A.P $=\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)d\right]$

$$\begin{gathered} =\frac{n}{2}\left[6+(n-1)2\right] \\ =\frac{n}{2}\left(4+2n\right) \\ =n\left(2+n\right) \end{gathered}$$

Step 2. Now, Take $\left(5+8+11+....10\right)$

It is also an A.P with $a=5$ and $d=3$

$\begin{array}{rcl}{S}_{10}& =& \frac{10}{2}\left(10+9\times 3\right)\\ & =& 5\times 37\end{array}$

Step 3. Put the value of $S_n$ and $S_{10}$ in given equation:

$\frac{3+5+7+....n}{5+8+11+....10}=7$

$\Rightarrow$ $\frac{n(2+n)}{5\times 37}=7$

$\Rightarrow$ $n(2+n)=7\times 5\times 37$

$\Rightarrow$ $n(n+2)=35\times 37$

$\mathbf{\therefore }\mathit{n}\mathbf{}\mathbf{=}\mathbf{}\mathbf{35}$

Hence, Option $\text{'}A\text{'}$ is Correct.