Question #2af10

May 23, 2017

Common ratio is $1 \frac{1}{2}$

Explanation:

$3$ rd term of A.P is $a + \left(3 - 1\right) b = a + 2 b$

$5$ th term of A.P is $a + \left(5 - 1\right) b = a + 4 b$

$8$ th term of A.P is $a + \left(8 - 1\right) b = a + 7 b$

$3$ consecutive terms of G.P are $a + 2 b , a + 4 b , a + 7 b$.
in G.P
$\frac{a + 4 b}{a + 2 b} = \frac{a + 7 b}{a + 4 b} \mathmr{and} {\left(a + 4 b\right)}^{2} = \left(a + 2 b\right) \left(a + 7 b\right)$ or

${\cancel{a}}^{2} + 8 a b + 16 {b}^{2} = {\cancel{a}}^{2} + 9 a b + 14 {b}^{2}$ or
$2 {b}^{2} = a b \mathmr{and} a = 2 b$

$3$ consecutive terms of G.P are $4 b , 6 b , 9 b$

Common ratio is $r = \frac{6 b}{4 b} = \frac{3}{2} \mathmr{and} \frac{9 b}{6 b} = \frac{3}{2}$ [Ans]