# finding the terms of a geometric progression

• April 1st 2012, 08:55 AM
pranay
finding the terms of a geometric progression
Hi,if we are given two equidistant terms of a geometric progression and also the sum of its terms , then how can we find the actual terms of the G.P? e.g if we are given the 3rd and last terms of the G.P as 12 and 48 and the sum of terms is 393 then the number of terms is 7 and they are: 3,6,12,24,48,96,192 .
I am aware of the following relations:
1. The sum of terms of G.P = a(r^n-1)/(r-1)
2. Product of equidistant terms of a G.P = product of extremes.
But am not sure how to apply them here to get the result.
Thanks.
• April 1st 2012, 09:10 AM
emakarov
Re: finding the terms of a geometric progression
Quote:

Originally Posted by pranay
if we are given the 3rd and last terms of the G.P as 12 and 48 and the sum of terms is 393 then the number of terms is 7 and they are: 3,6,12,24,48,96,192 .

How are 12 and 48 the "3rd and last" terms of the sequence? Isn't 192 the last term?

I am not sure what is given in your problem. I assume you are given that 12 are 48 are terms equidistant to the extremes and that the sum of terms is 393. Are you given the sequence length and the fact that 12 is the third term?
• April 1st 2012, 10:03 AM
biffboy
Re: finding the terms of a geometric progression
Can you make us a clear question here. Those terms add up to 381.
• April 1st 2012, 11:27 AM
Sylvia104
Re: Finding the terms of a geometric progression
Quote:

Originally Posted by pranay
Hi,if we are given two equidistant terms of a geometric progression and also the sum of its terms , then how can we find the actual terms of the G.P? e.g if we are given the 3rd and last terms of the G.P as 12 and 48 and the sum of terms is 393 then the number of terms is 7 and they are: 3,6,12,24,48,96,192 .

I think you are talking about the middle term of a GP with an odd number of terms. (The middle term is equidistant from the first and last terms (with respect to position in the progression) but we don't call it the "equidistant term"; that's bad English. (Shake)) By the way, if the middle term of a progression is the 3rd term, then the progression has 5 terms, not 7. Your GP is simply $3,6,12,24,48$ (and the sum of terms is $93$ (not 393)).

Quote:

Originally Posted by pranay
2. Product of equidistant terms of a G.P = product of extremes.

Better way to put it: Square of middle term of a GP = product of first and last terms.

Right, now that we've cleared things up, we can answer your question. The 3rd and 5th terms of a GP are $12$ and $48$ respectively, and the sum of the 5 terms is $93.$ Find the GP.

Solution:
Let the first term be $a.$ Using the fact that the square of the middle term is equal to the product of first and last terms, $12^2=144=48a$ $\implies$ $a=3.$ To find the common ratio, let it be $r.$ Equating the 3rd term from formula, $3r^2=12$ $\implies$ $r=\pm2.$ But if $r=-2$ then, using the sum-of-terms formula, the sum of the GP would be $33$ and not $93.$ Hence $r=2$ and the GP is $3,6,12,24,48.$
• April 1st 2012, 05:48 PM
pranay
Re: finding the terms of a geometric progression
Quote:

Originally Posted by emakarov
How are 12 and 48 the "3rd and last" terms of the sequence? Isn't 192 the last term?

I am not sure what is given in your problem. I assume you are given that 12 are 48 are terms equidistant to the extremes and that the sum of terms is 393. Are you given the sequence length and the fact that 12 is the third term?

i am extremely sorry ..given are the 3rd and 3rd last terms of the g.p(equidistant from extremes) .Here 12 and 48 . and the sum of terms is 381.