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

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

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

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

Cheers.

Alternative, if you use the geometric series formula.

S = \frac{a(1-r^n)}{1-r}

where a = first term, r = multiple factor, n = # of terms.

Hi Bunuel, how are these two formula different? Thank you.

The formula I used is for the sum of infinite geometric progression with common ratio |r|<1.

Hi All,

As complex as this question looks, it's got a great pattern-matching 'shortcut' built into it. When combined with the answer choices, you can avoid some of the calculations....

By plugging in the first few numbers (1, 2, 3, 4), you can see that a pattern emerges among the terms....

1st term = 1/2
2nd term = -1/4
3rd term = 1/8
4th term = -1/16

The terms follow a positive-negative-positive-negative pattern all the way to the 10th term and each term is the product of the prior term and 1/2. By "pairing up' the terms, another pattern emerges....

1/2 - 1/4 = 1/4

1/8 - 1/16 = 1/16

1/32 - 1/64 = 1/64

Etc.

The "pairs" get progressively smaller (notice how each is the product of the prior term and 1/4). This means that we're "starting with" 1/4 and adding progressively TINIER fractions to it. Since we're just adding 4 progressively smaller fractions to 1/4, this means that we're going to end up with a total that's just a LITTLE MORE than 1/4. Looking at the answer choices, there's only one answer that fits:

Final Answer:
GMAT assassins aren't born, they're made,
Rich

I've never seen this term "geometric progression" in my studies thus far - is there a good overview of them somewhere and potential questions that might be asked in reference to them? Thanks!

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

Hi healthjunkie,

Geometric progressions are rather rare on the GMAT (while you will see at least 1 sequence question on Test Day, it is not likely to be a Geometric sequence), so you shouldn't be putting too much effort into this concept just yet.

How are you performing on the Quant section overall? How about in the 'big' categories (Algebra, Arithmetic, Number Properties, DS, etc.)? That's where you're going to find the bulk of the points.

GMAT assassins aren't born, they're made,
Rich

I used the same calculation as above, which will probably take little more than 2 minutes. Is there a simple version to solve this problem?

Hi vijaydoli,

This question comes up every so often in this Forum. There are a couple of different ways of thinking about this problem, but they all require a certain degree of "math."

Without too much effort, you can deduce what the sequence is:

+1/2, -1/4, +1/8, -1/16, etc.

The "key" to solving this question quickly is to think about the terms in "sets of 2"…

1/2 - 1/4 = 1/4

Since the first term in each "set of 2" is greater than the second (negative) term, we now know that each set of 2 will be positive.

1/8 - 1/16 = 1/16

Now we know that each additional set of 2 will be significantly smaller than the prior set of 2.

1/4....1/16....1/64....etc.

Without doing all of the calculations, we know….
We have 1/4 and we'll be adding tinier and tinier fractions to it. Since there are only 10 terms in the sequence, there are only 5 sets of 2, so we won't be adding much to 1/4. Based on the answer choices, only one answer makes any sense…

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Sum of n terms of a GP = a(1 - r^n)/(1 - r)

The formula is the same whether |r| is more than 1 or less than 1.

You can find the sum of an infinite GP by the formula a/(1 - r) only when |r| < 1.
You cannot find the sum of an infinite GP when |r| > 1 because the sum will be infinite.
e.g. 3 + 9 + 27 + 81 ...... infinite terms - The sum will be infinite since you keep adding larger and larger terms.

Seriously - would anyone be able to resolve in 2 minutes?

Hi Erjan_S,

If you were trying to calculate the exact sum of this sequence, then you would likely find it almost impossible to do that in under 2 minutes. Thankfully, this question doesn't actually ask you to do that - the answer choices are all RANGES, which is a big 'hint' that you're supposed to something OTHER than calculate the exact sum. The 'key' to this question is to look at the sequence in 'pairs' (re: the 1st and 2nd, the 3rd and 4th, the 5th and 6th, etc.). Defining how pairs of terms relate to one another makes solving this question a lot easier than trying to calculate the sum of all 10 terms (my solution explains all of this in detail).

GMAT assassins aren't born, they're made,
Rich

Hi Bunuel,

Can we say that if |r| <1, then it's an infinite GP ? How does one define infinite GP?
When does one use the forumula

Sum = b1 (r^n-1)/r-1 ?

Please explain..

Infinite progressions are those with infinite number of terms. Whereas a finite sequence has defined first and last terms.

To add to what Bunuel said, r is the common ratio and has nothing to do with whether a progression is infinite or finite. In either case, r can be less than 1 or more than 1

e.g.
1, 3, 9, 27, 81 ... (infinite sequence with r = 3)
27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
1, 3, 9 (finite sequence with 3 elements and r = 3)
27, 9, 3 (finite sequence with 3 elements and r = 1/3)

Now the point is that the sum of all terms of the first sequence is infinite. The terms will can getting larger and will keep adding up. So the sum will be infinite.

We can find the exact sum of the rest of the 3 sequences.

27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
Sum = a/(1 - r) = 27/(1 - 1/3) = 81/2

1, 3, 9 (finite sequence with 3 elements and r = 3)
Sum = a(r^n - 1)/(r - 1) = 1*(3^3 - 1)/(3 - 1) = 13

27, 9, 3 (finite sequence with 3 elements and r = 1/3)
Sum = a(1 - r^n)/(1 - r) = 27*(1 - 1/3^3)/(1 - 1/3) = 26*3/2 = 39

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 an exact value. 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 out 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 1/2 and -1/4 are the largest term and the smallest term in our set, respectively, the sum will never fall below 1/4 or exceed 1/2.

Thus, we conclude that T is greater than 1/4 but less than 1/2.

Answer: D