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The 22nd Term

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by DanaJ » Tue Apr 14, 2009 3:40 pm
There's trusty formulas for this one.

I. 1 + 2 + 3 + ... + n = n(n+1)/2

II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.

Now apply that to n = 22:

A will be 22*23/2 = 11*23

The other number you're looking for will be 22*23*45/6 = 11*23*45/3 = 11*23*15 = 15 A.
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by gmat740 » Tue Apr 14, 2009 7:07 pm
I. 1 + 2 + 3 + ... + n = n(n+1)/2

II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.
I would like to proceed from where DanaJ left

II. [n(n+1)/2 ]*[(2n+1)/3]

= A*[(2n+1)/3]

put n= 22

A*[45/3]

=A*15 or 15A

Hope this helps

Karan
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by dtweah » Tue Apr 14, 2009 7:52 pm
gmat740 wrote:
I. 1 + 2 + 3 + ... + n = n(n+1)/2

II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.
I would like to proceed from where DanaJ left

II. [n(n+1)/2 ]*[(2n+1)/3]

= A*[(2n+1)/3]

put n= 22

A*[45/3]

=A*15 or 15A

Hope this helps

Karan
Voila!! Question designed to explore the link between the two formulas.
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by cramya » Tue Apr 14, 2009 8:03 pm
i would go with D

a^2

each term is squared

first series 22nd term A

second series 22nd term A^2
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by dumb.doofus » Tue Apr 14, 2009 9:47 pm
Yes.. the question just asks for the 22nd term.. which if it is A in the first sequence , then it would be A^2 in the second sequence..
answer seems to be D. or am I missing something..
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by deepoe » Thu Apr 16, 2009 2:30 am
Can't u use the formula A + (n-1)D ?
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by krisraam » Thu Apr 16, 2009 6:51 pm
deepoe wrote:Can't u use the formula A + (n-1)D ?
You can use that formula only if its a arithmetic progresssion.

Thanks
Raama
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