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)

$\frac{1}{2},\frac{1}{h_1},\frac{1}{h_2},\dots ,\frac{1}{h_{20}},\frac{1}{6}$ are in A.P.
$\Rightarrow \frac{1}{6}=\frac{1}{2}+21d\Rightarrow \frac{\frac{1}{6}-\frac{1}{2}}{21}=d$
$\Rightarrow d=-\frac{1}{63}$
$\Rightarrow \frac{1}{h_{18}}=\frac{1}{2}+18d=\frac{1}{2}+182\left(-\frac{1}{637}\right)=\frac{1}{2}-\frac{2}{7}$
$\frac{1}{h_{18}}=\frac{3}{14}\Rightarrow h_{18}=\frac{144}{3}$
Also, $2,a_1,a_2,\dots ,a_{20},6$ are in $A.P.$ $\Rightarrow 6=2+21D\Rightarrow D=\frac{4}{21}$
$\Rightarrow a_3=2+3D=2+3\times \frac{4}{247}=\frac{18}{7}$
$\Rightarrow a_3{h}_{18}=\frac{186}{y}\times \frac{142}{3}=12$