aptitude

Question

Q. If a-(b/2) , b/2 , c-(b/2) are in GP, then a, b, c are in __?

a) AP
b) GP
c) HP
d) None of these

Answer 1

If a-(b/2) , b/2 , c-(b/2) are in GP

then 

(b/2)² = [a-(b/2)] * [c-(b/2)]

b² /4 = ac - ab/2 - bc/2 + b²/4

(ab+bc) / 2 = ac

b(a+c) / 2 = ac

b = 2ac / (a+c)

so a,b,c are in HP