Three numbers a, b, c, none zero, form an arithmetic progression. Increasing a by 1 or increasing c by 2 results in a geometric progression. Then b equals:

1963 AMC 12 Problem 16

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16.

Three numbers

a,

b,

c,

none zero, form an arithmetic progression. Increasing

a

1

or increasing

c

2

results in a geometric progression. Then

b

equals:

16

14

12

10

8

Translate the three progression conditions into equations involving

b^2

and

a+c

Compare

(a+1)c=b^2=a(c+2)

first

The conditions give

\begin{gathered} 2b=a+c,\\ b^2=(a+1)c,\qquad b^2=a(c+2). \end{gathered}

Comparing the last two yields

c=2a.

Thus

b=\frac{3a}{2},

and

\frac{9a^2}{4}=2a(a+1).

Since

a\ne0,

a=8,

b=12.

Therefore, the correct answer is C.