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Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t

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Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t [#permalink]

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Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such that A has an even number of terms. If the sum of terms at odd positions is p and sum of terms at even positions is q, what is the ratio of p to q (take the common ratio as r)?

(A) r
(B) r2
(C) 1/r2
(D) 1/r
(E) None of these
[Reveal] Spoiler: OA

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Re: Latest Grail question bank PS 700 level [#permalink]

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New post 12 Apr 2013, 01:37
i guess GP over here means Geometric Progression....
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Re: Latest Grail question bank PS 700 level [#permalink]

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New post 12 Apr 2013, 01:49
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I think it should be 1/r

Let A be the GP with first term as a , common ratio as r, and number of terms as 6

GP = a, ar, ar^2, ar^3, ar^4, ar^5

Sum of GP = \(\frac{a (r^n - 1)}{r-1}\) where a=first term, n=number of terms, r=common ratio

Odd GP = a, ar^2, ar^4 Sum odd GP = \(\frac{a (r^6 - 1)}{r^2-1}\)

Even GP = ar, ar^3, ar^5 Sum even GP = \(\frac{ar (r^6 - 1)}{r^2-1}\)

Desired Ratio = \(\frac{a}{ar}\) = \(\frac{1}{r}\)

PS :- G.P. means Geometric Progression
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Re: Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such [#permalink]

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New post 12 Apr 2013, 04:56
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You dont need to know the formula for the Sum of GP to solve this

Just write the odd and even digits of the given statement together

P would become a +ar^2 + ar^4

Q would become ar + ar^3 + ar^5

=> P/Q = a(1 + r^2 + r^4) / ar (1 + r^2 + r^4)
=> P/Q = a(1 + r^2 + r^4) / ar(1 + r^2 + r^4)
=> P/Q = a/ar
=> P/Q = 1/r
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Re: Latest Grail question bank PS 700 level [#permalink]

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New post 27 Oct 2013, 17:01
Narenn wrote:
I think it should be 1/r

Let A be the GP with first term as a , common ratio as r, and number of terms as 6

GP = a, ar, ar^2, ar^3, ar^4, ar^5

Sum of GP = \(\frac{a (r^n - 1)}{r-1}\) where a=first term, n=number of terms, r=common ratio

Odd GP = a, ar^2, ar^4 Sum odd GP = \(\frac{a (r^6 - 1)}{r^2-1}\)

Even GP = ar, ar^3, ar^5 Sum even GP = \(\frac{ar (r^6 - 1)}{r^2-1}\)

Desired Ratio = \(\frac{a}{ar}\) = \(\frac{1}{r}\)

PS :- G.P. means Geometric Progression


Hi, how can the number of terms be 6. It must be 3 right? Since for the even and odd terms it is just 3. Is that correct?

Please clarify!
Thanks
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Re: Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t [#permalink]

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New post 03 Nov 2013, 18:15
Zarrolou wrote:
Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such that A has an even number of terms. If the sum of terms at odd positions is p and sum of terms at even positions is q, what is the ratio of p to q (take the common ratio as r)?

(A) r
(B) r2
(C) 1/r2
(D) 1/r
(E) None of these


If A = (\(a, ar, ar^2, ar^4,\)…), then shouldn't the next terms be \(ar^8\) and \(ar^16\). or am I missing something.
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Re: Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t [#permalink]

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Re: Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t [#permalink]

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New post 06 Nov 2015, 21:34
Narenn wrote:
I think it should be 1/r

Let A be the GP with first term as a , common ratio as r, and number of terms as 6

GP = a, ar, ar^2, ar^3, ar^4, ar^5

Sum of GP = \(\frac{a (r^n - 1)}{r-1}\) where a=first term, n=number of terms, r=common ratio

Odd GP = a, ar^2, ar^4 Sum odd GP = \(\frac{a (r^6 - 1)}{r^2-1}\)

Even GP = ar, ar^3, ar^5 Sum even GP = \(\frac{ar (r^6 - 1)}{r^2-1}\)

Desired Ratio = \(\frac{a}{ar}\) = \(\frac{1}{r}\)

PS :- G.P. means Geometric Progression



how did the denominator change from r-1 to r^2-1 ?
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Re: Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t [#permalink]

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New post 14 Dec 2015, 21:17
Please correct the question.
A = {a, ar, ar2, ar4,…..} <--- This is not a GP.
A = {a, ar, ar2, ar3,…..} <--- This is.
Re: Let A be a G.P. defined by A = {a, ar, ar2, ar4,…..}, such t   [#permalink] 14 Dec 2015, 21:17
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