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  • Given a sequence of 4 numbers,first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal,then the last term is :<

Given a sequence of 4 numbers,first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal,then the last term is :

Option: 1

16


Option: 2

8


Option: 3

4


Option: 4

2


Answers (1)

best_answer

Let the 4 numbers are

\mathrm{a,ar,ar^2,a}

\\\mathrm{since\;ar,ar^2,a\;\;are\;in\;AP}\\a+ar=2ar^2\\1+r=2r^2\\r=1,\;-\frac{1}{2}

r = 1 not possible because the common difference is 6 in AP

So,

\text { ar }-a r^{2}=6 \Rightarrow a=8

Posted by

Kuldeep Maurya

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