There are four numbers of which the first three are in G.P. and the last three are in A.P. with common difference 9. If the first and last number are equal, the sum of the four numbers is

There are four numbers of which the first three are in G.P. and the last three are in A.P.
with common difference 9. If the first and last number are equal,
the sum of the four numbers is

A
18
B
21
C
24
D
27

Solution:

As the common difference is 9, we let the four numbers be

$a, a - 18, a - 9, a$

As $a, a - 18, a - 9$ are in G.P.

$\begin{array}{l} (a - 18)^{2} = a (a - 9) \\ \Rightarrow a^{2} - 36 a + 324 = a^{2} - 9 a \\ \Rightarrow 27 a = 324 \Rightarrow a = 12 \end{array}$

Thus, sum of the four numbers is 21

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