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Question

The first three of four given numbers are in G.P. and last three are in A.P. whose common difference is 6. If the first and last numbers are same, then first will be?

A
2
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B
4
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C
6
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D
8
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Solution

The correct option is D 8
Last 3 of the 4 numbers are in AP.

Let they are, ad,a,a+d. Also the first number is same as 4th,a+d.

Therefore the 4 numbers are a+d,ad,a,a+d

The first 3 of these are in G.P.

(ad)2=a(a+d) But d=6

(a6)2=a(a+6)

a22a+36=0

Solving the above quadratic equation, we get,

a=2

Therefore the series is:

8,4,2,8


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