PS Question help

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 first 10 terms in the sequence, then T is

A. Greater than 2
B. Between 1 and 2
C. Between 1/2 to 1
D. Between 1/4 to1/2
E. Less than ¼

Calculate until you see the pattern.
Some test-takers might find it helpful to visualize the sum on a number line.

If k=1, -1^(1+1)*(1/2*1) = 1/2
If k=2, -1^(2+1)*(1/2*2) = -1/4
Sum of the first two terms is 1/2 + ( -1/4) = 1/4.

If k=3, -1^(3+1)*(1/2*3) = 1/8.
If k=4, -1^(4+1)*(1/2*4) = -1/16

Now we can see the pattern.
The sum increases by a fraction (1/8, for example) and then decreases by a fraction 1/2 the size (1/16).
In other words, the sum will alternate between increasing a little and then decreasing a little less than it went up.

The sum of the first 2 terms is 1/4. From there, the sum will increase by 1/8, decrease by a smaller fraction (1/16), increase by an even smaller fraction (1/32), and so on. Here are the first four terms, plotted on a number line:



Since all of the fractions after the first two terms will be less than 1/4, the sum will end up somewhere between 1/4 and 1/2.

The correct answer is D.

Sir
I have just started out 2 weeks ago. I was practicing OG13 PS but I have hit a wall. I am concentrating too much on speed rather than accuracy. These are my stats:
Q.1-30-> 29/30 in 32 min.
Q.31-60-> 26/30 in 19 min.
Q. 71-100-> 28/30 in 35 min.
Overall: Q.1-100-> 92/100 (just 92% accuracy on easy level questions)

Since the initial problems are supposedly easy, I am simply storming through the problems. Just cant stop myself. I am aiming over 770-790 and so I need your help.

Thank You

kkadvent wrote:For every integer k from 1 to 10 inclusive, the kth term of certain sequence is given by (-1)^(k+1) 1/2^k. If T is the sum of first 10 terms in sequence then T is

1)greater than 2
2)between 1 & 2
3)between 1/2 & 1
4)between 1/4 & 1/2
5)greater than 1/4

Lets create the progression using the values...

1/2 - 1/4 + 1/8 - 1/16..... 1/2^10

We can write it (1/4 + 1/16 + 1/64....)

Maximum SUM of the GP when there are infinite steps... 1/4(1/(1-1/4)) = 1/3 Formula used a/(1-r)... A is first term and r is ratio between 2 terms.
Minimum SUM of the GP when there is only one step = 1/4

[spoiler]So Answer lies between 1/4 to 1/3 or 1/4 to 1/2... [/spoiler]

Brent@GMATPrepNow wrote:

kkadvent wrote:For every integer k from 1 to 10 inclusive, the kth term of certain sequence is given by (-1)^(k+1) 1/2^k. If T is the sum of first 10 terms in sequence then T is

1)greater than 2
2)between 1 & 2
3)between 1/2 & 1
4)between 1/4 & 1/2
5)greater than 1/4

One way to solve this is to apply the formula for what's known as a geometric series, but I'm not really a fan of memorizing formulas.
Another option is to get a better idea of this sum, by finding a few terms:

T = 1/2 - 1/4 + 1/8 - 1/16 + . . .
We can rewrite this as T = (1/2 - 1/4) + (1/8 - 1/16) + . . .

When you start simplifying each part in brackets, you'll see a pattern emerge. We get...
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)
Note: 1/16 + 1/16 + 1/16 + 1/16 = 1/4
So, we can conclude that 1/16 + 1/64 + 1/256 + 1/1024 = (a number less 1/4)

Now start from the beginning: T = 1/4 + (1/16 + 1/64 + 1/256 + 1/1024) = 1/4 + (a number less 1/4)
= A number less than 1/2
Of course, we can also see that T > 1/4
So, [spoiler]1/4 < T < 1/2[/spoiler]

Cheers,
Brent

Hello Brent,

Thanks a lot for the excellent explanation. I was clear till this point:

Of course, we can also see that T > 1/4

Is there a way we can arrive at this result similar to how you have concluded that T = A number less than 1/2

Thanks for your help.

Best Regards,
Sri