Question

If $\frac{2}{3},H_1,H_2,H_3,H_4,\frac{2}{13}$ is an H.P., then the value of $\frac{H_1{H}_4}{H_2{H}_3}$ is

• A. $\frac{14}{27}$
• B. $\frac{21}{39}$
• C. $\frac{55}{42}$
• D. $\frac{63}{55}$

Answer 1

The correct option is D $\frac{63}{55}$

$\frac{2}{3},H_1,H_2,H_3,H_4,\frac{2}{13}$ is an H.P.
Then $\frac{3}{2},\frac{1}{H_1},\frac{1}{H_2},\frac{1}{H_3},\frac{1}{H_4},\frac{13}{2}$ is an A.P.
Let the common difference be $d$.
$\frac{13}{2}=\frac{3}{2}+5d\Rightarrow d=1$

Now,
$\frac{1}{H_1}=\frac{3}{2}+1=\frac{5}{2}\Rightarrow H_1=\frac{2}{5}$
$\frac{1}{H_2}=\frac{3}{2}+2=\frac{7}{2}\Rightarrow H_2=\frac{2}{7}$
$\frac{1}{H_3}=\frac{3}{2}+3=\frac{9}{2}\Rightarrow H_3=\frac{2}{9}$
$\frac{1}{H_4}=\frac{3}{2}+4=\frac{11}{2}\Rightarrow H_4=\frac{2}{11}$
Therefore, $\frac{H_1{H}_4}{H_2{H}_3}=\frac{\frac{4}{55}}{\frac{4}{63}}=\frac{63}{55}$