Question

if a b c are in GP show that (a²+b²), (ab+bc), (b²+c²) are in GP​

Answer 1

Step-by-step explanation:

if a, b, c are in GP, then

    b/a = c/b

or, b² =ac

now, if we substitute b² with ac in given terms we get

(a²+b²) = a²+ac = a(a+c).....term (i)

(ab+bc) =, b(a+c)..... term (ii)

(b²+c²) =ac+c² = c(a+c).....term (iii)

now ,

term (ii)² = [b(a+c)]²= b²(a+c)²

term (i) x term (iii) = a(a+c)*c(a+c) = ac*(a+c)² = b²(a+c)²

∴ term (ii)² = term (i) x term (iii)    [ hence proved ]