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f an infinite GP is summable (|r|<1) then the sum is \frac{b_1}{1-r}

can someone please explain what these means with numbers?

i guess what i'm asking for here is a question where we would use this concept

The sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.

For example, the sum of infinite geometric progression 1/2, 1/6, 1/18, 1/54, ... (first term = 1/2, common ratio = 1/3) is \(sum=\frac{b}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{3}}=\frac{3}{4}\).

"Summation The sum of an infinite GP will be finite if absolute value of r < 1 The general sum of a n term GP with common ratio r is given by \(b_1*\frac{r^n - 1}{r-1}\) If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\) "

Awesome post! After 4 years, I still want to read your post. Just need to correct the typo I hightlight than your post is perfect. The correct formula is \(S=b_1\frac{1-r^n}{1-r}\) not

"Summation The sum of an infinite GP will be finite if absolute value of r < 1 The general sum of a n term GP with common ratio r is given by \(b_1*\frac{r^n - 1}{r-1}\) If an infinite GP is summable (|r|<1) then the sum is \(\frac{b_1}{1-r}\) "

Awesome post! After 4 years, I still want to read your post. Just need to correct the typo I hightlight than your post is perfect. The correct formula is \(S=b_1\frac{1-r^n}{1-r}\) not

\(b_1*\frac{r^n - 1}{r-1}\)

There is no typo there. Those two are the same: factor -1 from denominator and numerator and reduce.
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One small correction as highlighted in bold: In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is ODD. In either case this is also equal to the mean of the first and last terms first and last terms

In "summation" for the arithmetic progression, where you give the general sum of a n term AP with common difference d, you have "a" instead of "a1", right?

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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..."n a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is odd. In either case this is also equal to the mean of the first and last terms"

Misc Notes A subsequence (any set of consequutive terms) of an AP is an AP

A subsequence (any set of consequutive terms) of a GP is a GP

A subsequence (any set of consequutive terms) of a HP is a HP

If given an AP, and I pick out a subsequence from that AP, consisting of the terms \(a_{i1},a_{i2},a_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be an AP

For Example : Consider the AP with \(a_1=1, d=2\) {1,3,5,7,9,11,...}, so a_n=1+2*(n-1)=2n-1 Pick out the subsequence of terms \(a_5,a_{10},a_{15},...\) New sequence is {9,19,29,...} which is an AP with \(a_1=9\) and \(d=10\) If given a GP, and I pick out a subsequence from that GP, consisting of the terms \(b_{i1},b_{i2},b_{i3},...\) such that \(i1,i2,i3\) are in AP then the new subsequence will also be a GP

For Example : Consider the GP with \(b_1=1, r=2\) {1,2,4,8,16,32,...}, so b_n=2^(n-1) Pick out the subsequence of terms \(b_2,b_4,b_6,...\) New sequence is {4,16,64,...} which is a GP with \(b_1=4\) and \(r=4\)

The special sequence in which each term is the sum of previous two terms is known as the fibonacci sequence. It is neither an AP nor a GP. The first two terms are 1. {1,1,2,3,5,8,13,...}

In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is even. In either case this is also equal to the mean of the first and last terms

I am sure you mean that " In a finite AP, the mean of all the terms is equal to the mean of the middle two terms if n is even and the middle term if n is ODD. In either case this is also equal to the mean of the first and last terms "

Thank you so much for the post though, is very valuable

Very helpful post indeed. I have studied lots of concepts using different prep sources and I must say what I learn on here on the forum is priceless. Sometimes I feel like what the heck did I study or how did I miss that such an elegant solution exists. Many thanks!
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If you must err, err on the side of hope. I believe in your success

Minor correction: It should be "a finite AP, the mean of all the terms is also equal to the mean of the middle two terms if n is even and the middle term if n is odd."

Minor correction: It should be "a finite AP, the mean of all the terms is also equal to the mean of the middle two terms if n is even and the middle term if n is odd."

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