M17

First of all: when we have geometric progression with common ratio \(q\) in the range \(-1<q<1\) \((|q|<1)\), then the sum of the progression: \(b_1, b_2, ...b_n...b_{+infinity}\) is \(Sum=\frac{b_1}{1-q}\).

In our case \(b_1=\frac{1}{2}\) and \(q=\frac{1}{2}<1\). The sum of the sequence \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}+...+\frac{1}{2^{infinity}}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1\). Which means that the sum of sequence given, even if it'll continue endlessly, will NEVER be more than 1 (actually it'll be equal to 1 as we calculated).

We have n=20, which is big enough to conclude that the sum will be very close to 1, but again never more than 1.

Answer: D.

Another way:

We have geometric progression with:
\(b_1=\frac{1}{2}\), \(q=\frac{1}{2}\), \(n=20\).

\(Sum=\frac{b_1(1-q^n)}{1-q}\)

\(S_{20}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{1-\frac{1}{2}}=\frac{\frac{1}{2}(1-\frac{1}{2^{20}})}{\frac{1}{2}}=1-\frac{1}{2^{20}}\)

Now: \(\frac{1}{2^{20}}\) is less than \(\frac{1}{10}\). Why? \(\frac{1}{2^4}=\frac{1}{16}<\frac{1}{10}\), so if \(\frac{1}{2^4}\) is less than \(\frac{1}{10}\), \(\frac{1}{2^{20}}\) will be much less than \(\frac{1}{10}\).

Next if we subtract the value less than \(\frac{1}{10}\) from \(1\), we'll get the value more than \(\frac{9}{10}\), as \(1-\frac{1}{10}=\frac{9}{10}\)

Hence \(1-\frac{1}{2^{20}}\) is more than 9/10 and clearly less than \(1\). The sum is between 9/10 and 1, only answer choice covering this range is D.

Answer: D.

Hope it's clear.

Same approach got D.

Thanks, that is very clear. Kudos!

this will be a GP series with a[1st term] = 1/2 and r[common ratio] =1/2 so
Sum of this series will be a(1-r^n)/(1-r) = 1/2 (1-[1/2]^20]/(1-1/2)= 1 - 1/(2^20)

from the given options option 4 is the best fit

Fact 1 : Sum of infinite series (1/2) + (1/2)^2 + (1/2)^3 + ..... to infinity is 1. (Using infinite GP formula). So option E is out.

Next thing is how does this sum look like ?
1 term : 0.5
2 terms : 1/2 + 1/4 = 0.75
3 terms : 1/2 + 1/4 + 1/8 = 7/8 = 0.875
... and so on

Basically with every term, the distance to 1, is halfed.

Option A can't be right either as it is bounded by 2/3, which is lower than the first 2 terms
Option B can't be right either as it is bounded by 3/4, which is lower than the first 3 terms

Choosing between C & D
This is where it gets interesting
As I mentioned, with each term, the distance to 1 is halfed.
After 2 terms we are 1/4 away
After 3 terms we are 1/8 away
...
After 20 terms we will be 1/(2^20 ) away

This is a very small number, much smaller than 0.1 (which is the bound implied by C)

Hence, the answer must be D

You need to know the formula for a geometric series:


\(SUM(ar^k) = \frac{a(r-r^{n+1})}{1-r}\) where k = m to n

For this series a = 1, k = \(\frac{1}{2}\) and n=20

you'll get the formula \(\frac{\frac{1}{2}-\frac{1}{2^{21}}}{\frac{1}{2}}\), approximate \(\frac{1}{2^{20}}\) with zero and you'll get 1. So you know that the series converges towards 1,

Add the first 4 terms to get 15/16 which is bigger than 9/10

This leaves answer D as the correct answer (I think )

Oops! my bad for posting the already asked question. I searched for it but didn't find in results. I think, I was searching the word s03, which showed only s03#1 (& s03#3) and didn't think of searching for progression (silly me). I think, introducing the new tag "Source:GMAT Club Tests" would be great.


Also, I did apply GP the way you guys mentioned but I think I didn't get the question very well. Is this question asking about where does the sum of the terms in question exist between given answer options? Am I getting it right?

Yes, it is asking for the range in which the sum sits amongst the given ranges

is there any other way (such as worst/best scenario) to solve this question?