Question

If $2,h_{1},h_{2},.....,h_{20},6$ are in harmonic progression and $2,a_{1},a_{2},....,a_{20},6$ are in arithmetic progression, then the value of $a_{3}h_{18}$ is equal to

• A. $6$
• B. $12$
• C. $3$
• D. $9$

Answer 1

Answer: (B)

$\frac{1}{2}, \frac{1}{h_{1}}, \frac{1}{h_{2}}, \ldots, \frac{1}{h_{20}}, \frac{1}{6}$ are in A.P.
$\Rightarrow \frac{1}{6}=\frac{1}{2}+21 d \Rightarrow \frac{\frac{1}{6}-\frac{1}{2}}{21}=d$
$\Rightarrow d=-\frac{1}{63}$
$\Rightarrow \frac{1}{h_{18}}=\frac{1}{2}+18 d=\frac{1}{2}+182\left(-\frac{1}{637}\right)=\frac{1}{2}-\frac{2}{7}$
$\frac{1}{h_{18}}=\frac{3}{14} \Rightarrow h_{18}=\frac{144}{3}$
Also, $2, a_{1}, a_{2}, \ldots, a_{20}, 6$ are in $A . P .$ $\Rightarrow 6=2+21 D \Rightarrow D=\frac{4}{21}$
$\Rightarrow a_{3}=2+3 D=2+3 \times \frac{4}{247}=\frac{18}{7}$
$\Rightarrow a_{3} h_{18}=\frac{186}{y} \times \frac{142}{3}=12$