Question

If $\frac{(5+9+13+\dots .n)}{(7+9+11+..12)}=\frac{5}{12}$ , then $n$ equals

• A. $5$
• B. $6$
• C. $9$
• D. $12$

Answer 1

The correct option is B

$6$

Explanation for correct option:

Let $S_1=5+9+13+\dots .n$ terms

Here the first term $a=5$ and common difference is $4$

Now,

$S_1=\frac{n}{2}\left[2a+(n-1)d\right]$

$=\frac{n}{2}\left[10+(n-1)4\right]$

$$\begin{gathered} =\frac{n}{2}(6+4n) \\ =3n+2n^2 \end{gathered}$$

$S_2=7+9+11+\dots 12$terms

$$\begin{gathered} =\frac{12}{2}(14+11\times 2) \\ =6(14+22) \\ =216 \end{gathered}$$

Divide $S_1$ by $S_2$, we get

$\frac{S_1}{S_2}=\frac{5}{12}$

$\Rightarrow$ $\frac{(3n+2n^2)}{216}=\frac{5}{12}$

$\Rightarrow$ $2n^2+3n-90=0$

$\Rightarrow$$(2n+15)(n-6)=0$

$\Rightarrow$$n=\frac{-15}{2}$ or $n=6$

$\therefore$The number of terms can not be negative so the answer is $n=6$ .

Hence, the option (B) is correct .