Question

Insert 4 HM between $\frac{2}{3}and\frac{2}{13}$

• A. $H_1=\frac{2}{5}$$H_2=\frac{2}{7}$
  $H_3=\frac{2}{9}$
  $H_4=\frac{2}{11}$
  
• B. $H_1=\frac{3}{5}$$H_2=\frac{2}{7}$$H_3=\frac{3}{9}$$H_4=\frac{2}{11}$
• C. $H_1=\frac{2}{5}$$H_2=\frac{3}{7}$$H_3=\frac{2}{9}$$H_4=\frac{3}{11}$
• D. $H_1=\frac{3}{5}$$H_2=\frac{2}{7}$$H_3=\frac{2}{9}$$H_4=\frac{3}{11}$

Answer 1

The correct option is A $H_1=\frac{2}{5}$

$H_2=\frac{2}{7}$

$H_3=\frac{2}{9}$

$H_4=\frac{2}{11}$

let the for HM inserted are $H_1,H_2,H_3,H_4$

so reciprocal of these numbers will make AP

common difference of AP will be:

$d=\frac{\frac{13}{2}-\frac{3}{2}}{5}=1$

therefore $\frac{1}{H_1}=\frac{3}{2}+1$

$H_1=\frac{2}{5}$

$\frac{1}{H_2}=\frac{5}{2}+1$

$H_2=\frac{2}{7}$

$\frac{1}{H_3}=\frac{7}{2}+1$

$H_3=\frac{2}{9}$

$\frac{1}{H_4}=\frac{9}{2}+1$

$H_4=\frac{2}{11}$