Progressions-9 (FMS-2010)

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) = x² (or) (x-d)(x+d+2) = x²

=> (x-d+1)(x+d) = (x-d)(x+d+2) => x²-d²+x+d = x²-d²+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)