If 2nd, 3rd and 6th terms of an AP are the three consecutive terms of a GP then

Solution:

We have been given that $2^{\text {nd }} 3^{\text {rd }}$ and $6^{\text {th }}$ terms of an $A P$ are the three consecutive terms of a GP.

Let the three consecutive terms of the G.P. be a,ar,ar $^{2}$.

Where $a$ is the first consecutive term and $r$ is the common ratio.

$2^{\text {nd }}, 3^{\text {rd }}$ terms of the A.P. are a and ar respectively as per the question.

$\therefore$ The common difference of the A.P. $=\mathrm{ar}-\mathrm{a}$

And the sixth term of the A.P. $=a r^{2}$

Since the second term is a and the sixth term is $a r^{2}$ (In A.P.)

We use the formula:t $=a+(n-1) d$

$\therefore a r^{2}=a+4(a r-a) \ldots\left(\right.$ the difference between $2^{\text {nd }}$ and $6^{\text {th }}$ term is $\left.4(a r-a)\right)$

$\Rightarrow a r^{2}=a+4 a r-4 a$

$\Rightarrow a r^{2}+3 a-4 a r=0$

$\Rightarrow a\left(r^{2}-4 r+3\right)=0$

$\Rightarrow a(r-1)(r-3)=0$

Here, we have 3 possible options:

1) $\mathrm{a}=0$ which is not expected because all the terms of A.P. and G.P. will be 0 .

2) $r=1$, which is also not expected because all th terms would be equal to first term.

3)r $=3$, which is the required answer.

Ans: Common ratio = 3