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Question

Three numbers a,b,c, non-zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Then b equals :
  1. 16
  2. 10
  3. 14
  4. 12


Solution

The correct option is D 12
a,b,c form an AP.
2b=a+c
Increasing a by 1 or c by 2 results in a GP
b2=(a+1)c      ...(1)
and b2=a(c+2)            ...(2)
(a+1)c=a(c+2)ac+c=ac+2ac=2aNow,2b=a+2ab=3a2
Putting this in (1), we get
9a24=(a+1)2a9a4=2a+29a=8a+8a=8b=3a2=3×82=12

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