A) 2
B) 4
C) 6
D) 8
Correct Answer: D
Solution :
Let the numbers be \[\frac{a}{r},\ a,\ ar,\ 2ar-a\] ?..(i) Where first three numbers are in G.P. and last three are in A.P. Given that the common difference of A.P. is 6, so \[ar-a=6\] ?..(ii) Also given \[\frac{a}{r}=2ar-a\Rightarrow \frac{a}{r}=2\,(ar-a)+a\] \[\Rightarrow \] \[\frac{a}{r}=2(6)+a,\] from (ii) \[\Rightarrow \] \[\left( \frac{a}{r} \right)-a=12\]\[\Rightarrow \]\[a(1-r)=12r\]\[\Rightarrow \]\[r=-\frac{1}{2}\] From (i) we get, \[a\left[ \left( -\frac{1}{2} \right)-1 \right]=6\]or \[a=-4\] Required numbers from (i) are \[8,\ -4,\ 2,\ 8\] .You need to login to perform this action.
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