The sum of the terms of a geometric progression is 2047.

Question

The sum of the terms of a geometric progression is 2047. Find the common ratio.

(1) The first and last terms of the series are 1 and 1024 respectively.

(2) Last but one term of the series is 512.

BrentGMATPrepNow · Answer

In case there are people reading this question and worrying that they haven't learned about geometric progressions and common ratios, you need not worry. The GMAT does not expect you to be familiar with these terms. Likewise, you don't need to know the formula for the sum of a geometric series/sequence/progression.

Cheers,
Brent

stne · Answer

Summation of n terms of a G.P = sum = a (\frac{ 1- r^n}{1-r})

Where a is the first term and n is the number of terms whose sum is being taken and r is the common ratio.

2047 =  1 (\frac{ 1- r^n}{1-r})

Also nth term of a G.P.=  ar^{(n-1)}  where a is the first term and r is the common ratio.
So last term ( nth term) = 1024 =  ar^{(n-1)} 
We know a (first term) is 1
 1024=r^{(n-1)} 
 2^{10}  = 1024 here n-1= 10 so n=11
 4^5  = 1024 here n-1= 5 so n=6 
 32^2  = 1024 here n-1= 2 so n=3

Out of these only n=11 will give sum of 2047 hence r= 2 is the required common ratio.

Hope this helps !
Kudos will be appreciated !
Thank you.

IanStewart · Answer

As Brent says above, you don't need to know anything about "geometric series" for the GMAT - you don't even need to know what they are. So ignoring the other problems with the wording of the question, it's simply out of the scope of the test. The only way it could appear is if the question itself also provided the formulas you'd need to use to answer it.

But since the only solution above assumes r is a positive integer, and uses inspection, I can offer a different method, though test takers can safely ignore questions like this. We know from Statement 1 that the first term a = 1, and the last term is 1024. In a geometric sequence with n terms, the last term is equal to (a)(r^(n-1)). Since a =1, we know

r^(n-1) = 1024

The sum S of a geometric sequence with n terms is equal to

S = a(1- r^n)/(1 - r)

so since that sum is 2047, we also know, using that a=1,

2047 = (1 - r^n) / (1 - r)

Since r^n is just equal to (r)(r^(n-1)) by basic exponent rules, and since r^(n-1) = 1024, we can replace r^n with 1024r. So

2047 = (1 - 1024r) / (1 - r)
2047 - 2047r = 1 - 1024r
2046 = 1023r
2 = r

and Statement 1 is sufficient. Statement 2 is not sufficient (it could be the sequence with r=2 that we found in Statement 1, but since r need not be an integer, it could just be a two-term sequence where the first term is 512, and the second is 1535, among many possibilities), so the answer is A.

chandra73 · Answer

Thank you Brent for Clarification!

Samuel0709 · Answer

To solve this problem we just need to know the formula that defines the nth term in a G.P : 
An= A1 R^(n-1)
Given an=1024 and a1=1
=> 1024= 1 x R^(n-1)
=> 1024= R^(n-1)
=> for a given sum of the sequence we can find R. Hence Statement one is sufficient.
Statement two merely tells us the second last element in the sequence is 512. Without knowing the last element we cannot conclude on what the common ratio might be

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The sum of the terms of a geometric progression is 2047.

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