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Kth Term

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by MBALA2009 » Sun Jul 13, 2008 9:00 pm
For every integer k from 1 to 10, the kth term of a certain sequence is given by (-1)^k+1 * (1/2^k). If T is the sum of the first 10 terms in the sequence, then T is

A) greater than 2
B) between 1 and 2
C) between 1/2 and 1
D) between 1/4 and 1/2
E) less than 1/4

ANSWER = D
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by sudhir3127 » Sun Jul 13, 2008 10:05 pm
Even i got the answer as E .. but not sure if what i am doing is right or wrong. i usef Gometric mean to solve the equation.

Assumed K = 1 in (-1)^k+1 * (1/2^k).

i get the first term as 1/2 ,

take K=2 we get -1/4 ... K=3 we get 1/8..

hence in GM

sum of n terms in GM is a(1- r^n)/(1-r)

solving this usong a= 1/2 n=10 and r =-1/2

we get 341/1024 = .333


which lies between 1/2 and 1/4.

let me know if this approach is right.
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by Ian Stewart » Mon Jul 14, 2008 1:16 am
I suggested a couple of ways to solve this problem here:

www.beatthegmat.com/sequence-t12719.html

You can do the problem without knowing any formulas for summing geometric series, and without adding all of the terms.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by ildude02 » Mon Jul 14, 2008 10:52 am
Can you please explaing a little bitmore about your GM formulae. What kind of sequences you are supposed to use this and how do you get the "r" value?
sudhir3127 wrote:Even i got the answer as E .. but not sure if what i am doing is right or wrong. i usef Gometric mean to solve the equation.

Assumed K = 1 in (-1)^k+1 * (1/2^k).

i get the first term as 1/2 ,

take K=2 we get -1/4 ... K=3 we get 1/8..

hence in GM

sum of n terms in GM is a(1- r^n)/(1-r)

solving this usong a= 1/2 n=10 and r =-1/2

we get 341/1024 = .333


which lies between 1/2 and 1/4.

let me know if this approach is right.
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by BlueRain » Mon Jul 14, 2008 12:42 pm
geometric series is any sequence where the (n+1)th term is r * nth term. Thus r is the common ratio.

Example, 1, 2, 4, 8, 16, r would be 2.
In this particular problem, you see 1/2, -1/4, 1/8, -1/16, so the common ratio is -1/2 (notice the negative sign).

For the geometric series sum, assume S as the sum, and a as the first term
S=a+a*r+a*r^2+a*r^3+a*r^4+...+a*r^(n-1) (Eq. 1, n-1 because technically n starts at 0, a = a * r^0)
r*S=a*r+a*r^2+a*r^3+a*r^4+...+a*r^(n-1)+a*r^n (multiply Eq. 1 by r on both sides)
r*S - S = a*r^n - a (cancel all the common terms)
S(r-1) = a[r^n - 1] (factor out S on the left side and a on the right side)
S = a[r^n - 1]/(r-1)

If you take the negative of both numerator and the denominator, you would get the equation above given by sudhir.
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by ildude02 » Mon Jul 14, 2008 1:11 pm
Thanks a bunch, BlueRain. It makes sense now. One follow up question to make sure is, can the term "a" or the starting term of the series be any number or it needs to start with "1" ? I see that the sequence here starts with "1/2" as the a term, so may be we can say that "a" doesn't need to start with 1.
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by BlueRain » Mon Jul 14, 2008 2:00 pm
a is just the first term of the sequence, it does not have to be 1.
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