Question

If there is a sequence, in Hp (Harmonic
Progression), as 2/12,
a, b, c, 1/16 find a, b, c.​

Answer 1

Answer:

$a=\frac{2}{17}$

$b=\frac{1}{11}$

$c=\frac{2}{27}$

Step-by-step explanation:

Concept used:

A sequence of numbers is said to be in H.P. if their reciprocals are in A.P.

Given:

$\frac{2}{12},a,b,c,\frac{1}{16}$ are in H.P

$\implies\:\frac{12}{2},\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{16}{1}$ are in A.P

$\implies\:6,\frac{1}{a},\frac{1}{b},\frac{1}{c},16$ are in A.P

Then,

$\frac{1}{a}=6+d$

$\frac{1}{b}=6+2d$

$\frac{1}{c}=6+3d$

$16=6+4d$

$10=4d$

$d=\frac{10}{4}=\frac{5}{2}$

Now,

$\frac{1}{a}=6+\frac{5}{2}=\frac{17}{2}$

$\implies\:a=\frac{2}{17}$

$\frac{1}{b}=6+2(\frac{5}{2})=11$

$\implies\:b=\frac{1}{11}$

$\frac{1}{c}=6+3(\frac{5}{2})=\frac{27}{2}$

$\implies\:c=\frac{2}{27}$