Question

If $9$ A.M.s and $9$ H.M.s are inserted between $2$ and $3$ and $A$ be any A.M. and $H$ be the corresponding H.M.,

then $H(5-A)=$

• A. $10$
• B. $6$
• C. $-6$
• D. None of these

Answer 1

Answer: (B)

$6$

Let A be the $k^{th}$ A.M., then H will be the $k^{th}$ H.M.,
Now, $A=2+kd=2+k\left[\frac{3-2}{10}\right]=\frac{20+k}{10}$

$\frac{1}{H}=\frac{1}{2}+k{d}^{\prime }=\frac{1}{2}+k\left[\frac{\frac{1}{3}-\frac{1}{2}}{10}\right]=\frac{30-k}{60}$

$\therefore A+\frac{6}{H}=5⟹H(5-A)=6$
Hence, option 'B' is correct.