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Math Expert V
Joined: 02 Sep 2009
Posts: 58411
For every integer m from 1 to 10 inclusive, the mth term of a certain  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 65% (02:23) correct 35% (02:39) wrong based on 116 sessions

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For every integer m from 1 to 10 inclusive, the $$m_{th}$$ term of a certain sequence is given by $$(-1)^{(m+1)}*(\frac{1}{2})^m$$. 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 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25

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Math Expert V
Joined: 02 Sep 2009
Posts: 58411
Re: For every integer m from 1 to 10 inclusive, the mth term of a certain  [#permalink]

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Bunuel wrote:
For every integer m from 1 to 10 inclusive, the $$m_{th}$$ term of a certain sequence is given by $$(-1)^{(m+1)}*(\frac{1}{2})^m$$. 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 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25

_________________
Intern  B
Joined: 14 Mar 2018
Posts: 4
Re: For every integer m from 1 to 10 inclusive, the mth term of a certain  [#permalink]

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

so the answer is d between .25 and .5
Director  D
Joined: 24 Oct 2016
Posts: 535
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 Re: For every integer m from 1 to 10 inclusive, the mth term of a certain  [#permalink]

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1
Bunuel wrote:
For every integer m from 1 to 10 inclusive, the $$m_{th}$$ term of a certain sequence is given by $$(-1)^{(m+1)}*(\frac{1}{2})^m$$. 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 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25

TL;DR

b_1 = (-1)^2*(1/2)^1 = 1/2
b_2 = (-1)^3*(1/2)^2 = -1/4
b_3 = (-1)^4*(1/2)^3 = 1/8

b_1 = 1/2; r = -1/2

Sum of n term GP = b_1 * (1-r^n)/(1-r)
= (1/2) (1-(1/1024)/(1+(1/2))
= (1/2) ((1024-1)/1024)/(3/2) = 1023/(3*1024) = ~1/3

Veritas Prep Official Solution

We get the first term by putting m = 1

First term = [(-1)^(1+1)]*[(1/2)^1] = 1/2 (Not so bad, eh?!)

Second term = [(-1)^(2+1)]*[(1/2)^2] = -1/2^2 = -1/4

Third term = [(-1)^(3+1)]*[(1/2)^3] = 1/2^3 = 1/8

Do we see a pattern?

The tenth term will be [(-1)^(10+1)]*[(1/2)^10] = -1/2^10 = -1/1024 (You don’t need to calculate this of course. You can keep it as 2^10. I do suggest though that you should be good with the first 10 powers of 2, first 6 powers of 3, first 4 powers of 4 and first 3 powers of 5 to 10.)

The sequence looks like this: 1/2, – 1/4, 1/8, – 1/16, …

T = 1/2 – 1/4 + 1/8 – 1/16 +…+ 1/512 – 1/1024

Of course GMAT doesn’t expect us to calculate this. One could end up wasting a lot of precious time if one did. The trick is to know that we need to figure out the answer using some shrewdness. Fortunately (or unfortunately), GMAT software rewards cunning and craft!

The problem is that we have positive and negative terms so it is very hard to say what the value of T will be. But the terms are not random. There is a positive term followed by a negative term which is then followed by a positive term and so on. Also, every subsequent term is smaller than the previous term (in fact it is a Geometric Progression but we don’t need to know that to solve this question. Nevertheless, we will take up GP too in a couple of weeks).

We need to create some uniformity so that we can deduce something about T. 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. The first number is greater than the second number in each group e.g. 1/2 is greater than 1/4, therefore, (1/2 – 1/4) = 1/4 i.e. a positive number
Similarly, (1/8 – 1/16) = 1/16, again a positive number.

This means T = 1/4 + 1/16 +…. (all positives)
Definitely this sum, T, is greater than 1/4 i.e. 0.25

So we can rule out option (E). But we still have to choose one out of the four remaining options.

Now, let’s group the terms in another way.

T = 1/2 + (- 1/4 + 1/8) + (- 1/16 + 1/32) … – 1/1024

You leave out the first term and start grouping two terms at a time. The last term will be left alone too! You will be able to make four groups with the 8 terms in the middle.

Now look closely at each group: The first term is a negative number with a higher absolute value while the second term is a smaller positive number so the sum will give you a negative number, e.g.:

(-1/4 + 1/8) = -1/8

(-1/16 + 1/32) = -1/32 etc

This means T = 1/2 – 1/8 – 1/32 … -1/1024

All 4 of the groups will give you a negative number and the last term is also negative. 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 has to be option (D).
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Math Expert V
Joined: 02 Aug 2009
Posts: 8004
Re: For every integer m from 1 to 10 inclusive, the mth term of a certain  [#permalink]

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Bunuel wrote:
For every integer m from 1 to 10 inclusive, the $$m_{th}$$ term of a certain sequence is given by $$(-1)^{(m+1)}*(\frac{1}{2})^m$$. 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 0.5 and 1
(D) between 0.25 and 0.5
(E) less than 0.25

the $$m_{th}$$ term of a certain sequence is given by $$(-1)^{(m+1)}*(\frac{1}{2})^m$$ tells us that it is a geometric progression, where r=$$-\frac{1}{2}$$ and first term is $$m_1=\frac{1}{2}$$

MAX value

So the sum of first 10 terms will be less than the sum of infinite sequence, and sum of infinite sequence is $$\frac{m_1}{1-r}=\frac{1}{2}/(1-(\frac{-1}{2})=\frac{1}{2}/\frac{3}{2}=\frac{1*2}{2*3}=\frac{1}{3}=0.333$$

MIN value

Now what is the minimum value first term is positive next becomes half but negative, so each pair will give us a positive term $$(m_1-m_2)+(m_3-m_4)...$$
So, let us just check only $$(m_1-m_2)=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}=0.25$$
So answer is between 0.25 and 0.33.

D
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CrackVerbal Quant Expert G
Joined: 12 Apr 2019
Posts: 268
Re: For every integer m from 1 to 10 inclusive, the mth term of a certain  [#permalink]

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This question is a fairly easy one on sequences. Using simple mathematical logic, at least 2 of the 5 options can be eliminated. This then, is the way to go.

First, let us try to calculate the initial few terms of the sequence and try and see if we can get a pattern.

The mth term is defined as $$(-1)^{m+1}$$ * $$(\frac{1}{2})^m$$ for all values of m between 1 to 10.

Plugging in the value of m as 1, we get, 1st term = ½. Similarly, 2nd term = -$$\frac{1}{4}$$, 3rd term = $$\frac{1}{8}$$ and 4th term = $$\frac{1}{16}$$.

So, the sequence looks like this: ½ - ¼ + $$\frac{1}{8}$$– $$\frac{1}{16}$$……..

The biggest number in this sequence is ½. From ½, we are subtracting some values. Therefore, the sum cannot be greater than ½ (or 0.5). So, options A, B and C can be eliminated straight away. Possible answers are D or E.

If we look at the first 4 terms, we see that the total value of these terms is ¼ + $$\frac{1}{16}$$, effectively. This is more than ¼ (or 0.25). Also, in this sequence, we are always subtracting a smaller number from a larger number. Therefore, the effective sum will be positive; you don’t have to worry that ¼ + $$\frac{1}{16}$$ will suddenly do a U-turn and end up becoming much smaller – that can’t happen.

Therefore, option E can now be eliminated since we know that the sum will be greater than 0.25. So, the correct answer option HAS to be D.

If you follow this approach, solving this question can be completed within a minute’s time.

Hope this helps!
_________________ Re: For every integer m from 1 to 10 inclusive, the mth term of a certain   [#permalink] 08 Jun 2019, 10:59
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