1) AP
2) HP
3) GP
4) None of these
Solution: (1) AP
Since a, b, c are in GP, b2 = ac
ax2 + 2bx + c = 0
ax2 + 2 √ac x + c = 0
(√ax + √c)2 = 0
x = – √c / a
Since ax2 + 2bx + c = 0 and dx2 + 2ex + f = 0 have common root.
x = – √c / a must satisfy dx2 + 2ex + f = 0
d * (c / a) – 2e (√c / a) + f = 0
(d / a) – (2e / √ac) + (f / c) = 0
(2e / b) = (d / a) + (f / c)
(d / a), (e / b), (f / c) are in AP.