JEE Main 2016 — Sequences And Series Question with Solution

The terms of an Arithmetic Progression (A.P.) are given by $a$, $a+d$, $a+2d$, ..., where $a$ is the first term and $d$ is the common difference.

Given that the 2nd, 5th and 9th terms of an A.P. are in Geometric Progression (G.P.), we can denote them as follows :

2nd term = $a+d$

5th term = $a+4d$

9th term = $a+8d$

For three numbers to be in G.P., the square of the middle term must be equal to the product of the other two terms. So,

$(a+4d{)}^2=(a+d)(a+8d)$

Expanding and simplifying :

$a^2+8ad+16d^2=a^2+9ad+8d^2$

$8ad+16d^2=a^2+9ad+8d^2$

$8ad-9ad=8d^2-16d^2$

$-ad=-8d^2$

$a=8d$

The common ratio of the G.P. is the ratio of the 5th term to the 2nd term, or $(a+4d)/(a+d)$. Substituting $a=8d$ gives :

$(8d+4d)/(8d+d)=12d/9d=4/3$

So, the common ratio of the G.P. is $4/3$. The answer is option D.