Q.If \(a, b, c\) are in continued geometric progression, prove that \(\cfrac{1}{b} = \cfrac{1}{b - a} + \cfrac{1}{b - c}\).

If \(a, b, c\) are in continued geometric progression, then \(\cfrac{a}{b} = \cfrac{b}{c}\) Let \(\cfrac{a}{b} = \cfrac{b}{c} = k\) where \(k \ne 0\) \(\therefore a = bk = ck^2,\) and \(b = ck\) Left-hand side = \(\cfrac{1}{b} = \cfrac{1}{ck}\) Right-hand side = \(\cfrac{1}{b - a} + \cfrac{1}{b - c}\) \(= \cfrac{1}{ck - ck^2} + \cfrac{1}{ck - c}\) \(= \cfrac{1}{ck(1 - k)} + \cfrac{1}{c(k - 1)}\) \(= \cfrac{1}{ck(1 - k)} - \cfrac{1}{c(1 - k)}\) \(= \cfrac{1 - k}{ck(1 - k)} = \cfrac{1}{ck}\) \(\therefore\) Left-hand side = Right-hand side (Proved)
Similar Questions