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

Question

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

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

Narenn · Accepted Answer

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

nikhilsehgal · Answer

i guess GP over here means Geometric Progression....

aceacharya · Answer

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

shankar245 · Answer

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

suk1234 · Answer

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.

GMATDemiGod · Answer

how did the denominator change from r-1 to r^2-1 ?

dprabhak2810 · Answer

Please correct the question.
 A = {a, ar, ar2, ar4,…..} <--- This is not a GP.
 A = {a, ar, ar2, ar3,…..} <--- This is.

mrfrantic · Answer

Ans (D)

The given GP series is as follows: { a ,  ar ,  ar^2 ,  ar^3 ,  ar^4 ,...}

Elements at odd position are  {a, ar^2, ar^4,...}

So, sum of elements at odd position,  p  =  a  +  ar^2  +  ar^4  + ... =  a*(1 + r^2 + r^4 + ... )

or,  \frac{p}{a}  =  (1 + r^2 + r^4 + ... )        ...(1)

Now, elements at even position are  {ar, ar^3, ar^5,...}

And their sum,  q  =  ar + ar^3 + ar^5 + ...  =  ar*(1 + r^2 + r^4 + ... )

Using (1),  q  =  ar*\frac{p}{a}

or,  \frac{p}{q} = \frac{1}{r}

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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

Prep Toolkit

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