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[yvichman]

For every integer K from 1 to 10, inclusive, 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 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

[/spoiler]

[DanaJ]

Let's say q = -1/2. Then T would be:
1/2 (q^0+q^1+q^2+...+q^9). That "thing" between the parenthesis has a formula you can use to calculate it.:
q^0+q^1+...+q^n = [q^(n+1)-1]/q-1. So T would be:

1/2{[q^10-1]/q-1}. Since q = -1/2 and is smaller that 1, we can safely say T =1/2{ [1-(-1/2)^10]/[1+1/2]} = [1 - (1/2)^10]/3. That means T is 1/3 minus smth very small (1/1024*3), so T is close to 1/3.
Since 1/4 < 1/3 <1/2, I'd go with D. Please tell me if I'm right, I'm not sure the formula above is accurate (it's been quite some time since high school...).

[Brent@GMATPrepNow]

For every integer K from 1 to 10, inclusive, 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 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
[/spoiler]

Dana is right in that you can use a formula to find the sum of the terms of a geometric sequence. However, we can also tackle it a different way.
First, get an idea of this sum. Find the first few terms to get:
T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .
Or T = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024
Now examine the last 4 terms ( 1/16 + 1/64 + 1/256 + 1/1024)
Notice that 1/64, 1/256, and 1/1024 are each less than 1/16
So, (1/16 + 1/64 + 1/256 + 1/1024) < (1/16 + 1/16 + 1/16 + 1/16) = 1/4
Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024), but since 1/16 + 1/64 + 1/256 + 1/1024 is less than 1/4 we know that T is less than 1/4 + 1/4 (1/2)
So, 1/4 < T < 1/2 (D)