Question

Insert 2 no.s between 1 & $\frac{1}{3}$ so that the sequence becomes an Harmonic progression --

• A. $\frac{5}{3},\frac{7}{3}$
• B. $\frac{5}{7},1$
• C. $\frac{3}{5},\frac{3}{7}$
• D. $\frac{1}{7},\frac{2}{7}$

Answer 1

The correct option is C $\frac{3}{5},\frac{3}{7}$

Inserting 2 no.s in H.P is equivalent to inserting reciprocals of them in A.P.
So. 1, 3 are in A.P. and we have to insert two more no. in such a way that it becomes an A.P.
We know, $1+\frac{b-a}{n+1}$

First inserted term = $1+\frac{3-1}{2+1}=1+\frac{2}{3}=\frac{5}{3}$
Second inserted term = $1+2\left(\frac{2}{3}\right)=1+\frac{4}{3}=2+\frac{1}{3}=\frac{7}{3}$
So, terms will be - $\frac{3}{5},\frac{3}{7}$