# If the second, fourth and ninth term of an arithmetic progression are in geometric progression, then what is the common ratio of the GP?

##### 1 Answer
Dec 4, 2017

Please refer to the Discussion in The Explanation.

#### Explanation:

Let the AP be, $a , a + d , a + 2 d , \ldots , a + \left(n - 1\right) d , \ldots \left(n \in \mathbb{N} .\right) .$

Then, by what is given, $a + d , \left(a + 3 d\right) \mathmr{and} \left(a + 8 d\right)$ are in GP.

$\therefore {\left(a + 3 d\right)}^{2} = \left(a + d\right) \left(a + 8 d\right) .$

$\therefore {a}^{2} + 6 a d + 9 {d}^{2} = {a}^{2} + 9 a d + 8 {d}^{2} , \mathmr{and} ,$

${d}^{2} - 3 a d = 0 , i . e . ,$

$d \left(d - 3 a\right) = 0.$

$\therefore d = 0 , \mathmr{and} , d = 3 a .$

If $d = 0 ,$ then, the GP becomes the constant sequence

$a , a , a , \ldots a , \ldots$ for which the common ratio is $1. \left(a \ne 0\right)$

In case, $d = 3 a ,$ then the GP is $4 a , 10 a , 25 a$ for which the

common ratio is $2.5 a , \mathmr{and} , \frac{5}{2} a , \left(a \ne 0.\right)$