There are four numbers of which the first three are in GP and the...

There are four numbers of which the first three are in GP and the last three are in AP, whose common difference is 6. If the first and the last numbers are equal, then two other numbers are

Options

A-2,4
B-4,2
C2,6
DNone of the above

Correct answer

B. -4,2

Step-by-step solution

Let 3 numbers in AP are

a,(a+6),(a+12)

Also, first and last number out of 4 numbers are equal

\therefore 4

numbers are

(a+12), a,(a+6),(a+12)

Also, given, first 3 numbers are in GP

\begin{array}{ll} \Rightarrow & a^2=(a+12)(a+6) \\ \Rightarrow & a^2=a^2+18 a+72 \\ \Rightarrow & a=-\frac{72}{18}=-4 \end{array}

\therefore 4

numbers are

8,-4,2,8$

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