Question

if a b c are in gp then prove that 1/a+b 1/2b 1/b+c are in ap

Answer 1

Proved that $\frac{1}{a + b}, \frac{1}{2b}, \frac{1}{b + c}$ are in A.P.

Step-by-step explanation:

Given that a, b, and c are in G.P.

Therefore, ac = b² ............ (1)

We have to prove that $\frac{1}{a + b}, \frac{1}{2b}, \frac{1}{b + c}$ are in A.P.

So, we have to prove that, $\frac{1}{a + b} + \frac{1}{b + c} = 2(\frac{1}{2b}) = \frac{1}{b}$.

Now, $\frac{1}{a + b} + \frac{1}{b + c}$

= $\frac{a + b + b + c}{(a + b)(b + c)}$

= $\frac{a + c + 2b}{ab + ac + b^{2} + bc }$

= $\frac{a + c + 2b}{ab + b^{2} + b^{2} + bc }$ {Since, ac = b², from equation (1)}

= $\frac{a + c + 2b}{ab +2b^{2} + bc }$

= $\frac{a + c + 2b}{b(a + c + 2b) }$

= $\frac{1}{b}$

Hence, proved that $\frac{1}{a + b}, \frac{1}{2b}, \frac{1}{b + c}$ are in A.P. (Answer)