Three non-zero numbers a,b,c form an arithmetic progression.
Increasing a by 1 or increasing c by 2 results in a geometric progression. Then
b equals:

(1)16 (2)14 (3)12 (4)10

Solution follows here:

__Solution:__
As a,b,c are in AP, let them be x-d,x,x+d respectively.

**“**Increasing a by 1 or increasing c by 2 results in a geometric progression

**”**

**=>**x-d+1,x,x+d (or) x-d,x,x+d+2 are in G.P

=> (x-d+1)(x+d) = x

^{2}(or) (x-d)(x+d+2) = x^{2}
=> (x-d+1)(x+d) = (x-d)(x+d+2) => x

^{2}-d^{2}+x+d = x^{2}-d^{2}+2x-2d
=> x = 3d

Then x-d,x,x+d become 2d,3d,4d

=> given 2d+1,3d,4d are in G.P. => (3d)

^{2}= (2d+1)(4d)
on solving, d = 4 => the numbers are 8,12,16 => b=12

**Answer (3)**
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