​ If a, b, c are in GP, prove that 1/(a+b), 1/(2b), 1/(b + c) are in AP. ​

Question

If a, b, c are in GP, prove that \(\frac{1}{(a+b)}, \frac{1}{(2b)}, \frac{{1}}{(b+c)}\) are in AP.

Answer 1

To prove: \(\frac{1}{(a+b)}, \frac{1}{(2b)}, \frac{{1}}{(b+c)}\) are in AP

Given : a, b, c are in GP

Formula used: When a,b,c are in GP, b² = ac 

When a,b,c are in GP, b² = ac

Taking \(\frac{1}{(a+b)},\,and \,, \frac{{1}}{(b+c)}\)

\(\frac{1}{(a+b)} +\frac{{1}}{(b+c)}\)

\(\Rightarrow\) \(\frac{b + c + a +b}{(a + b)(b +c)}\)

\(\Rightarrow\) \(\frac{a + c+ 2b}{ab + ac + b^2 + bc}\)

\(\Rightarrow\) \(\frac{a + c + 2b}{ab + b^2 + b^2 + bc}\) [b² = ac]

\(\Rightarrow\) \(\frac{a + c + 2b}{ab + 2b^2 + bc}\)

\(\Rightarrow\) \(\frac{a + c + 2b}{b(a + 2b + c)}\)

\(\Rightarrow\) \(\frac{1}{b}\)

\(\Rightarrow\) \(2 \times \frac{1}{2b} \)

We can see that  \(\frac{1}{(a+b)} +\frac{{1}}{(b+c)}\) = \(2 \times \frac{1}{2b} \)

Hence we can say that \(\frac{1}{(a+b)}, \frac{1}{(2b)}, \frac{{1}}{(b+c)}\) are in AP.