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 :

• A. 16
• B. 14
• C. 12
• D. 10

Answer 1

The correct option is C 12

a,b,c form an AP.
$\therefore 2b=a+c$
Increasing a by 1 or c by 2 results in a GP
$\therefore b^2=(a+1)c$ ...(1)
$and{b}^2=a(c+2)$ ...(2)
$\therefore (a+1)c=a(c+2)\therefore ac+c=ac+2a\therefore c=2aNow,2b=a+2a\therefore b=\frac{3a}{2}$
Putting this in (1), we get
$\frac{9a^2}{4}=(a+1)2a\therefore \frac{9a}{4}=2a+2\therefore 9a=8a+8\therefore a=8\therefore b=\frac{3a}{2}=\frac{3\times 8}{2}=12$