Question

If $4H{M}^{\prime }s$ be inserted between $2/3$ and $2/13$, then the second $HM$ is

• A. $2/5$
• B. $2/7$
• C. $2/11$
• D. $2/17$

Answer 1

Answer: (B)

$2/7$

Let $4H{M}^{\prime }$ be $a.b,c,d$

then $\frac{2}{3},a,b,c,d,\frac{2}{13}$ are in $H.P$

$\Rightarrow \frac{3}{2},\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d},\frac{13}{2}$ are in $AP$

with $a=\frac{3}{2}$ $\&$ $a6=\frac{13}{2}$

$a6=\frac{13}{2}$

$\Rightarrow \frac{3}{2}+5d=\frac{13}{2}$

$\Rightarrow 5d=\frac{13}{2}-\frac{3}{2}=5$

$\Rightarrow d=1$

Now,$a_3=a+2d$

$\Rightarrow \frac{1}{b}=\frac{3}{2}+2\times 1$

$\Rightarrow \frac{1}{b}=\frac{3+4}{2}$

$\Rightarrow \frac{1}{b}=\frac{7}{2}$

$\Rightarrow b=\frac{2}{7}$

Hence second $H.M$ is $\frac{2}{7}$