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shrouded1
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gurpreetsingh
Great initiative.

Suggestions :

1. Include AM, GM , HM included between 2 numbers.
2. use a_{n} with math tag to get \(a_{n}\)
3. Include more examples.

This post,along with the algebra, is a good initiative to fill the important topics of Math Book that were not included earlier.

Check
Check
Check

Let me know what else ?
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Great initiative. +1 to you.
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Great one :)
Kudos to you !
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added a couple more solved GMAT style questions to the end
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Good to know. It can definitely same valuable time.
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This is great post but I was just wondering whether we need to know these concepts for GMAT
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There are several questions in the OG that use these concepts. So I think its good to know all this. Plus if you search through the forums you'll find several Qs on sequences and progressions as well
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Valuable resource.
Kudos +1 :)
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I had a question. I am not sure if this works: "In case of n numbers : AM * HM = GM^n".

This seems to work for 2 numbers but for more than 2, it seems to break, please let me know if I am missing something.
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shrouded1
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nitantsharma
I had a question. I am not sure if this works: "In case of n numbers : AM * HM = GM^n".

This seems to work for 2 numbers but for more than 2, it seems to break, please let me know if I am missing something.

You are correct, this should only hold for special case n=2. Thanks for pointing out
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How can we get the formular for sumation of GEOMETRIC PROGRESSION. Please, prove, so that I do not have to remember the formular but to know the way to get the formular and so can solve the relative questions.
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!
This proof is beyond the scope of the GMAT

The proof below is based on mathematical induction

To prove : The sum of an n term GP : \(b,br,br^2,...,br^{n-1}\) is \(b*\frac{r^n-1}{r-1}\)

P(1 term) : The sum of the GP {b} is \(b*\frac{r^1-1}{r-1}=b\). Which is true trivially

P(n terms) : Let the sum of an n term GP : \(b,br,...,br^{n-1}\) be \(b*\frac{r^n-1}{r-1}\)

P(n+1 terms) : Consider the n+1 term GP : \(b,br,....,br^n\)
Sum of this GP = Sum of n term GP + \(br^n\) = \(b*\frac{r^n-1}{r-1} + br^n\)
Sum = \(\frac{b}{r-1} * (r^n - 1 + r^n(r-1))\)
=\(\frac{b}{r-1} *(r^{n+1}-1)\)

Hence P(1) is true
And if we assume P(n) true P(n+1) is true
By mathematical induction P(k) must be true for all k>=1
Hence, proved
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Can someone please clarify:

I think the formula for calculating the sum of n consecutive numbers should be:

\((lastterm - firstterm)*(lastterm - firstterm + 1)/2\)
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nonameee
Can someone please clarify:

I think the formula for calculating the sum of n consecutive numbers should be:

\((lastterm - firstterm)*(lastterm - firstterm + 1)/2\)

1,2,3,4,5

let's apply your formula on the above series.

\(\frac{(5-1)*(5-1+1)}{2} = \frac{4*5}{2} = 10\)

this is not correct. let's try another formula.

\(\frac{n}{2}(firstterm + lastterm)\)

\(\frac{5}{2}(5+1) = 15\)


\(\frac{1}{2}(firstterm + lastterm)\) basically gives you the avg of the series. when you multiply the avg with number of terms (n), you get the sum.

HTH
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dimitri92, thanks. I must have made a computational error.
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i know i read it somewhere, but i cant find it now. What are the ways to get the full Math book?
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The general sum of a n term GP with common ratio r is given by

For sum of a GP:
When r > 1, the denominator is (r-1)
When r < 1, the denominator is (1-r)
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