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.
The 22nd Term
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gmat740
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I would like to proceed from where DanaJ leftI. 1 + 2 + 3 + ... + n = n(n+1)/2
II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.
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.gmat740 wrote:I would like to proceed from where DanaJ leftI. 1 + 2 + 3 + ... + n = n(n+1)/2
II. 1^2 + 2^2 + 3^2 + ... + n^2 = n(n+1)(2n+1)/6.
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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dumb.doofus
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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..
answer seems to be D. or am I missing something..
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