For every integer k from 1 to 10, inclusive the "kth term of a certain

To get a hang of what the question is asking, put values for k right away. Say, k = 1, k = 2 etc
You get terms such as (1/2) when k = 1, (-1/4) when k = 2 etc

T = 1/2 - 1/4 + 1/8 - 1/16 +.... + 1/512 - 1/1024 (Sum of first 10 terms)

Of course GMAT doesn't expect us to calculate but figure out the answer using some shrewdness.

We have 10 terms. If we couple them up, two terms each, we get 5 groups:
T = (1/2 - 1/4) + (1/8 - 1/16) ...+ (1/512 - 1/1024)

Tell me, can we say that each group is positive? From a larger number, you are subtracting a smaller number in each bracket. e.g. 1/2 is larger than 1/4 so 1/2 - 1/4 = 1/4 i.e. a positive number
1/8 - 1/16 = 1/16, again a positive number.

We will get something similar to this: T = 1/4 + 1/16 +.... (all positives)
Definitely this sum, T, is greater than 1/4 i.e. 0.25

Now, let's group them in another way.

T = 1/2 + (- 1/4 + 1/8) + (- 1/16 + 1/32) ... - 1/1024
You will be able to make 4 groups since you left the first term out. The last term will also be left out.
Each bracket will give you a negative term -1/4 + 1/8 = -1/8 etc
Since the first term is 1/2 i.e. 0.5, we can say that the sum T will be less than 0.5 since all the other terms are negative.

So the sum, T, must be more than 0.25 but less than 0.5

Answer (D)

Check the basics of sequences here: https://anaprep.com/algebra-introducing-sequences/
and another tricky question here: https://anaprep.com/algebra-a-difficult ... sequences/

Initially, I thought about calculating every term. This is almost always the wrong approach. It is more of a number properties problem than sequences.

For the first term, it alternates between positive and negative. For even k, it is positive 1 * \(1/2^k\) and negative 1 for odd k.

The first term is \(1 * 1/2 = 1/2\)

The second term is \(-1 * 1/4 = -1/4\)

\(1/2 - 1/4 = 1/4\)

The third term is \(1 * 1/8 = 1/8\)

\(1/4 + 1/8 = 3/8\)

Looking at the answer choices, you don't need to continue. Since the denominator is increasing exponentially, the terms added and subtracted are becoming closer to 0. From the first term, we know we will never go above \(1/2\). After subtracting the second term, we know we will never go below \(1/4\).

Hello!

I was just doing this problem and though I would add some "graphic" approach in case it is useful for someone as it has been for me.

You can easily draw a graph of the first points on any series to see if they follow a regular pattern. This one specifically is extremely easy to catch as you draw 3+ points, something like this:

A few things here:

* I've never seen a real GMAT question that requires one to know any geometric sequence formulas. Of course there are questions where you might use such formulas, but there will always be a different approach available;

* In any sequence question which gives an expression for each term, you'll always want to write down the first few terms using the given expression to work out what the sequence looks like. Here we have:

\(\frac{1}{2}, \frac{-1}{4}, \frac{1}{8}, \frac{-1}{16}, \frac{1}{32}, \frac{-1}{64}, \frac{1}{128}, \frac{-1}{256}, \frac{1}{512}, \frac{-1}{1024}\)

* Now, adding all of these fractions together would take a long time to do. The GMAT *never* requires you to perform any crazy calculations, so there must be a different way to answer the question. Notice the answer choices are only estimates, so we only need to estimate the sum of the first 10 terms. When we want to estimate the value of a sum, we ignore terms that make only a tiny contribution to the sum. The last few terms in our sequence are minuscule compared to the first few, so to get a good estimate, we can completely ignore them; adding, say, the first four terms will give a perfectly good approximation of the sum here (you get 5/16, which is enough to choose the right answer).

* The sequence in this question is what is known as an 'alternating sequence' -- that is, the terms alternate between positive and negative values. When adding an alternating sequence, you most often want to add your terms in pairs first, grouping one positive and one negative (add the 1st and 2nd term, the 3rd and 4th, and so on). One doesn't need to do this for this question, but it does make the answer a bit easier to see - we'd find our sum is

\(\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}\)

from which you can instantly see the sum is greater than 1/4. Since every term here is tiny after the first two, the sum is certainly less than 1/2.

Kth term: A(k) = (-1) ^ (k+1) * 1 / 2^k

T is the sum of the first 10 terms of this sequence.

A(1) = (-1)^(1+1)*1/2^1 = 1/2

A(2) = (-1)^(1+2)*1/2^2 = -1/4

You can see that for odd k it would be positive values of \(1/2^k\) and for even k it would be negative values of \(1/2^k\)

So it's like this:

\(T = \frac{1}{2} - \frac{1}{4} + \frac{1}{18} - \frac{1}{16}+ ...\)

Now look for a pattern in the sums of various number of terms.

i.e. Sum of first 2 terms = 1/2 - 1/4 = 1/4
Sum of first 3 terms = 1/4 + 1/8 = 3/8
Sum of first 4 terms = 3/8 - 1/16 = 5/16
Sum of first 5 terms = 5/16 + 1/32 = 11/32
and so on...

You can see that all these sums are between 1/4 and 1/2.

On the GMAT when you have tested such a question for about half the terms and you have established a pattern you can safely assume it will continue. This is because this series is analogous to how the GMAT itself adapts the difficulty of its questions based on your answers. It sort of zigzags in a diminishing pattern. So as you increase k, you will be varying only slightly around a sort of ultimate stagnant value at infinity which you could determine as \(\frac{a}{(1 - r)}\) i.e. the sum of an infinite Geometric Progression with first term a and common ratio r.

Here we have first term 1/2 and common ratio -1/2

=> Infinite sum = \(\frac{1}{2*(1 - (-1/2))}= \frac{1}{2*1.5} = \frac{1}{3}\)

With a large number of terms such as 10 terms, your series sum would actual TEND to stagnate around this infinite sum.

Pick D

an alternative:
The series can be written as \(1/2 - 1/4 + 1/8 - 1/16.............1/1024\)
Since the terms beyond \(1/16\) are too small, so I am not considering those terms.
Now on adding \(1/2 - 1/4 +1/8 - 1/16\), we get \(5/16\) which is slightly more than \(4/16\). Hence its value would be around `\(0.3\).

The answer choice which includes this number is D.

When doing sequence problems, it usually helps to look at at least the first few terms. So in this case:

\(a_k = (-1)^{k+1} * \frac{1}{2^k}\)

This gives us:

\(a_1 = \frac{1}{2}\)
\(a_2 = -\frac{1}{4}\)
\(a_3 = \frac{1}{8}\)
\(a_4 = -\frac{1}{16}\)

We can stop there. The first, and largest, term is 1/2, and we then subtract 1/4. We will then add and subtract fractions that will continue to get smaller and smaller. So we can immediately eliminate A, B, and C, since the sum cannot possibly be greater than 1/2. Now we're left with D and E. Note that the sum of the first two terms is 1/4. Then you add another 1/8 and subtract 1/16. This pattern will continue all the way through the tenth term, and you should be able to see that there's no way this sum will become less than 1/4. So the answer is D.

Using some keen observation, you can quickly arrive at the answer...
Terms will be: \(\frac{1}{2} - \frac{1}{4} + \frac{1}{8} - \frac{1}{16} + \frac{1}{32} - ... - \frac{1}{1024}\)
For every pair of values:
\(\frac{1}{2} - \frac{1}{4} = \frac{1}{4}\)

\(\frac{1}{8} - \frac{1}{16} = \frac{1}{16}\)
etc...

So this series is actually just
\(\frac{1}{4} + \frac{1}{16} + ... + \frac{1}{1024}\)

So the sum is definitely greater than 1/4.
When you add an infinite GP with 1/16 as first term and 1/4 as common ratio, the sum will be \(\frac{\frac{1}{16}}{1-\frac{1}{4}} = 1/12\). Here, the sum of terms 1/16 + 1/64 + ... 1/1024 is definitely less than 1/12. Hence the sum is definitely less than 1/2. Answer is (D).

I did it the following way.

K=(1/2)-(1/4)+(1/8)-(1/16)+..... ----- 1

Multiply K by 2

2K=1-(1/2)+(1/4)-.....-(1/512) ------- 2

Adding 1 and 2

3K = 1 -(1/1024)
K= (1/3) -{1/(3*1024)}

Now 1/(3*1024) will be very small
So K= 1/3 = .3333

Ans Option D

This is GP.

The terms will be 1/2-1/4+1/8-....

Common ratio is (-1/4)/(1/2) = -1/2

So the sum of terms = 1/2 [1- (-1/2)^10]/(1-(-1/2)) = 1/2 *[1-1/1024]/3/2 = 1023/(1024*3) close to 1/3 so Option D

I liked Graphical approach is a good one.
BUT - felt there is no need of any calculations
T =
+(1/2 + 1/2^3 + 1/2^5 + 1/2^7 +1/2^9)
-(1/2^2 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10)
Subtract the top and bottom
T = 1/4 + 1/2^4 + 1/2^6 + 1/2^8 + 1/2^10
Now I know T is greater than 1/4 , but if sum of the remaining is smaller than 1/4, then it will definitely be between 1/4 and 1/2
T = 1/4 + 1/4(1/2^2 + 1/2^4 + 1/2^6 + 1/2^8)
So sum of the remaining is smaller than 1/4, so ans. is between 1/4 and 1/2
OA -D

Bunuel, how come the ratio here is -1/2, and not 1/2? If there are alternating signs in a sequence, is the "ratio" always the negative value? thanks

Solution:
We are given that for every integer k from 1 to 10, inclusive, the kth term of a certain sequence is given by (-1)^(k+1) x (1/2^k). We must determine the sum of the first 10 terms in the sequence. Before calculating the sum, we should recognize that the answer choices are provided as ranges of values, rather than as exact values. Thus, we might not need to calculate the total of the 10 terms to determine an answer. Perhaps we can uncover a pattern to help us find the answer. Let’s start by listing the first four terms.
k = 1:
(-1)^(1+1) x (1/2^1)
(-1)^2 x 1/2
1 x 1/2 = 1/2
k = 2:
(-1)^(2+1) x (1/2^2)
(-1)^3 x 1/4
-1 x 1/4 = -1/4
k = 3:
(-1)^(3+1) x (1/2^3)
(-1)^4 x 1/8
1 x 1/8 = 1/8
k = 4:
(-1)^(4+1) x (1/2^4)
(-1)^5 x 1/16
-1 x 1/16 = -1/16
Recall that we are trying to estimate the value of T = 1/2 + (-1/4) + 1/8 + (-1/16) + … until we have 10 terms. In other words, T = 1/2 – 1/4 + 1/8 – 1/16 + … until there are 10 terms.
We should notice that the absolute values of the terms are getting smaller:
|1/2|>|-1/4|>|1/8|>|-1/16|.
Notice that starting from the first term of 1/2, we are subtracting something less than 1/2 (notice that 1/4 < 1/2) but then adding back something even less (notice 1/8 < 1/4), and the process continues. Thus, because ½ and -1/4 are the largest term and the smallest term, respectively, in our set, the sum will never fall below ¼ or exceed ½.
Thus, we conclude that T is greater than 1/4 but less than 1/2.

Answer: D

The sum of infinite geometric progression with common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.

https://tinypic.com/r/2lvk4z5/8

Please check a simpler solution to the above problem in the above image link.

Cheers.

Here is a post that discusses Geometric progressions (GP):
https://anaprep.com/algebra-benefits-of ... -concepts/

Here we need to find a pattern
\(\frac{1}{2},-\frac{1}{4},\frac{1}{8},...\) as you see the sign changes every term.
The first and bigger is 0,5 and then we subtract and sum smaller and smaller terms.
We can eliminate any option that gives us a upper limit greater than 1/2.
We are down to D and E. Is the sum less than 1/4?
Take the sum of pair of terms : the first 2 give us \(\frac{1}{4}\), the second pair is \(\frac{1}{8}-\frac{1}{16}\) positive so we add value to \(\frac{1}{4}\), so the sum will be greater.(this is true also for the next pairs, so we add to \(\frac{1}{4}\) a positive value for each pair)
D

Hope its clear, let me know