1. ## AP/GP question

Here goes:

The first 3 terms of a geometric progression, with first term a = 1 and common ratio r, r does not equal 1, are the 1st, 2nd and 4th term of an arithmetic progression. Find the value of r. (Answer: r = 2)

I got 3 as my answer though. Thanks in advance if you could help solve this question!

2. Originally Posted by margaritas
Here goes:

The first 3 terms of a geometric progression, with first term a = 1 and common ratio r, r does not equal 1, are the 1st, 2nd and 4th term of an arithmetic progression. Find the value of r. (Answer: r = 2)

I got 3 as my answer though. Thanks in advance if you could help solve this question!
For the G.P.,
$a = 1$
$common$ $ratio = r (r$ $is$ $not$ $equal$ $to$ $1)$
$a_2=r$
$a_3=r^{2}$

Now since the first three terms of the G.P. are the first, second and fourth terms of an A.P., therefore for the A.P.
$t = 1$
$t_2=r$
$t_4=r^{2}$
$common$ $difference,d=t_2-t=r-1$

$Since$ $fourth$ $term$ $of$ $the$ $A.P.$ $is$ $the$ $third$ $term$ $of$ $the$ $G.P.$
$a_3=t_4=t+3d$
$r^{2}=1+3(r-1)$
$r^{2}=1+3r-3$
$r^{2}-3r+2=0$
$(r-2)(r-1)=0$
$r=1,r=2$
$But$ $since$ $r$ $is$ $not$ $equal$ $to$ $1(given),$ $hence$ $\boxed{r=2}.$
I got 3 as my answer though.
I think you probably made a mistake in factorisation.

3. Oh so the 'a' for AP is assumed to be 1 as well?

EDIT: Thanks for the solution, shubhadeep!

4. Originally Posted by margaritas
Oh so the 'a' for AP is assumed to be 1 as well?
Its not an assumption. Your question clearly states that.
Originally Posted by margaritas
The first 3 terms of a geometric progression, with first term a = 1 and common ratio r, r does not equal 1, are the 1st, 2nd and 4th term of an arithmetic progression.
The first term of the G.P. is the first term of the A.P., the second term of the G.P. is the second term of the A.P. and the third term of the G.P. is the fourth term of the A.P.

5. Oh ok, silly me.