If a,b,c are in A.P., then 2ax +1 , 2bx +1, 2cx +1 , x ≠ 0 are in

(1) A.P

(2) G.P only when x > 0

(3) G.P if x < 0

(4) G.P for all x ≠ 0

Solution:

Since a, b, c are in AP

2b = a+c

(2bx +1)2 = (22bx +2)

= 2(a+c)x +2

= 2ax+1 × 2cx+1

=> 2ax +1 , 2bx +1, 2cx +1 are in GP.

Hence option (4) is the answer.

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