Question

a,b,c are said to be in harmonic progression if the reciprocals 1/a,1/b,1/c are in arithmetic progression what will be the value of x for which 3,x,6 are in harmonic progression

Answer 1

Answer: 4

Step-by-step explanation:

Since, if 3, x, 6 are in Harmonic progression,

⇒ $\frac{1}{3}, \frac{1}{x} , \frac{1}{6}$ are in Arithmetic progression,

⇒ $\frac{1}{x} - \frac{1}{3} = \frac{1}{6} - \frac{1}{x}$

⇒ $-\frac{1}{3} - \frac{1}{6} = -\frac{1}{x} - \frac{1}{x}$

⇒ $\frac{-2-1}{6} = -\frac{2}{x}$

⇒ $-\frac{3}{6} = -\frac{2}{x}$

⇒ $\frac{1}{2} = \frac{2}{x}$

⇒ $x=4$

Answer 2

Answer:

The value of x is 4.

Step-by-step explanation:

We are given that,

$3,x,6$ are in harmonic progression.

Since, we know,

If $a,b,c$ are in harmonic progression, then $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in arithmetic progression.

Thus, we get,

$\frac{1}{3},\frac{1}{x},\frac{1}{6}$ are in arithmetic progression.

Further, we know,

If the terms are in arithmetic progression, then their common difference are equal.

So, we get,

$\frac{1}{x}-\frac{1}{3}=\frac{1}{6}-\frac{1}{x}$

i.e. $\frac{2}{x}=\frac{1}{6}+\frac{1}{3}$

i.e. $\frac{2}{x}=\frac{1+2}{6}$

i.e. $\frac{2}{x}=\frac{3}{6}$

i.e. $\frac{2}{x}=\frac{1}{2}$

i.e. x= 4

Hence, the value of x is 4.