# Given a sequence of four numbers such that first three terms are in G.P and the last three terms are in A.P with the common difference of 6. If the first and the fourth are equal then the common ratio of the G.P ?

Sep 27, 2017

The common ratio of G.P is $- \frac{1}{2}$

#### Explanation:

2nd,3rd and 4th number are in A.P with common difference $d = 6$

Let 2nd,3rd and 4th number are $x , x + 6 , x + 12$

1st and 4th number are equal , so 1st number is $x + 12$ , hence

four numbers of the sequence are $x + 12 , x , x + 6 , x + 12$

1st,2nd and 3rd number are in G.P, which means

$x + 12 , x , x + 6$ are in G.P . $\therefore \frac{x}{x + 12} = \frac{x + 6}{x}$

by cross multiplication , ${x}^{2} = \left(x + 12\right) \left(x + 6\right)$ or

${\cancel{x}}^{2} = {\cancel{x}}^{2} + 18 x + 72 \therefore 18 x = - 72 \mathmr{and} x = - 4$

$x + 12 = - 4 + 12 = 8 , x = - 4 , x + 6 = - 4 + 6 = 2 , x + 12 = 8$

So the numbers in the sequence are $8 , - 4 , 2 , 8$

The common ratio of G.P is $r = - \frac{4}{8} = \frac{2}{-} 4 = - \frac{1}{2}$ [Ans]