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Retired Moderator Joined: 29 Oct 2013
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The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is give by Sn = n/2(2a + (n-1)*d) = . What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.

A. 4345
B. 4302
C. 4258
D. 4214
E. 4170

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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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24
14
I think it is faster to ignore the given formula all together.

Lets use most commonly known formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms
Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050
Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836
Required sum : 5050 - 836 = 4214
Ans: D
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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12
3
We require to find:

(1 + 2 + 3 + 4........... + 99 + 100) - (26 + 28 + 30 + ............ + 60 + 62)

Expression First
(1 + 2 + 3 + 4........... + 99 + 100)

$$= \frac{100}{2} * (100 + 1) = 50 * 101 = 5050$$

Expression Second
(26 + 28 + 30 + ............ + 60 + 62)

$$Total terms = \frac{62-26}{2} + 1 = 18 + 1 = 19$$

$$Addition = \frac{19}{2} (26 + 62) = 44 * 19 = 836$$

Answer = 5050 - 836 = 4214
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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I don't think that on the test, I would know to change 63 and 25 to 62 and 26. If you try to find the number of terms using 63 and 25, you wind up with 20:

[(63 - 25)/2] + 1 = 20. And 22 times the average of 63 and 25 is 880, which if subtracted from 5,050 would lead you to answer choice D.

How do you recognize this trap and avoid falling into it?
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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for the 2nd part of the problem, omitting the even integers 25-63, why does the gmat explanation assign d=2?
I understand assigning a=26 and n=19 but not following why d=2.
Any assistance would be appreciated.

thanks
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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WishfulThinking1983 wrote:
I don't think that on the test, I would know to change 63 and 25 to 62 and 26. If you try to find the number of terms using 63 and 25, you wind up with 20:

[(63 - 25)/2] + 1 = 20. And 22 times the average of 63 and 25 is 880, which if subtracted from 5,050 would lead you to answer choice D.

How do you recognize this trap and avoid falling into it?

By recognizing the fact that we have to subtract the sum of EVEN numbers between 25 and 63 from the sum of all natural numbers from 1 to 100, inclusive.
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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2
2
jainaba wrote:
for the 2nd part of the problem, omitting the even integers 25-63, why does the gmat explanation assign d=2?
I understand assigning a=26 and n=19 but not following why d=2.
Any assistance would be appreciated.

thanks

The first part of the problem is straight forward:
Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050

In the second part, you need this sum

26 + 28 + 30 + ... + 62

This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2).
26 + (26+2) + (26+2*2) + .... (26+2*18)
a, a+d, a+2d,..., a+(n-1)d

So n-1 = 18 and hence, n = 19

Using the formula given, you get
Sum = n/2(2a + (n-1)*d) = 19/2 * [2*26 + (19 -1)*2] = 836

So the required sum = 5050 - 836 = 4214

For more on arithmetic progressions, check: http://www.veritasprep.com/blog/2012/03 ... gressions/
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GMAT 1: 760 Q50 V44 Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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MensaNumber wrote:
The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is give by Sn = (n/2)*(2a + n-1)d. What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.

A. 4345
B. 4302
C. 4258
D. 4214
E. 4170

There are two parts to this question:

1st part: Sum of integers from 1 to 100 inclusive
We can find this by applying the formula
Sn = (n/2)*(2a + n-1)d.

The other way to write this formula is (n/2)*(a + l), where a = first term and l = last term

Applying the formula,
Sn = (100/2)(1+100) = 50*101 = 5050

Second Part: Sum of even integers between 25 and 63
The even integers start from 26 and end at 62
No of terms = 19

Hence the sum = (19/2)*(26 + 62) = 836

Subtracting this from 5050, we get 4214. Option D
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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NoHalfMeasures wrote:
I think it is faster to ignore the given formula all together.

Lets use most commonly know formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms
Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050
Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836
Required sum : 5050 - 836 = 4214
Ans: D

where does the 19 come from in sum of evens? and why 62 and not 64?
Intern  Joined: 12 Nov 2015
Posts: 43
Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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Took 1.5 minutes. D is the correct one.

1-100 integer sum= 5050
25-63 even integer sum= 836

difference = 4214
Math Expert V
Joined: 02 Aug 2009
Posts: 7995
Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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1
2
NoHalfMeasures wrote:
The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is give by Sn = (n/2)*(2a + n-1)d. What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.

A. 4345
B. 4302
C. 4258
D. 4214
E. 4170

Hi,
a method to do it under one minutes is to take advantage of the choices given...
lets work on the last digit as we have different units digit in each choice...

total sum of 1 to 100 inclusive will have 0 as the last digit..
this is so because the resultant will be 10*(sum of all single digits)... and since we are multiplying by 10, units digit will be 0...

now for single digit in sum of even number from 25 to 63..
25 to 65 will have 4 times sum of single digit even int, 4*(2+4+6+8+0)=4*20..
here too the units digit is 0, but 64 has to be excluded from the total..

two ways from here on..

1) we are subtracting 0 from 0
so units digit should be 0, but we have to add 64..
so last/units digit =4..

2)we subtract 64 from sum of even int..
so units digit=80-4=76..
or units digit =6...

so our answer should have units digit as 10-6=4..
only D has 4 as units digit..
ans D

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GMAT 1: 640 Q39 V38 GMAT 2: 700 Q47 V38 Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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I got a question!

To find the # of even terms we have to divide by 2.

What if we have a ODD numbers of terms? What do we do?
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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VeritasPrepKarishma wrote:
jainaba wrote:
for the 2nd part of the problem, omitting the even integers 25-63, why does the gmat explanation assign d=2?
I understand assigning a=26 and n=19 but not following why d=2.
Any assistance would be appreciated.

thanks

The first part of the problem is straight forward:
Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050

In the second part, you need this sum

26 + 28 + 30 + ... + 62

This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2).
26 + (26+2) + (26+2*2) + .... (26+2*18)
a, a+d, a+2d,..., a+(n-1)d

So n-1 = 18 and hence, n = 19

Using the formula given, you get
Sum = n/2(2a + (n-1)*d) = 19/2 * [2*26 + (19 -1)*2] = 836

So the required sum = 5050 - 836 = 4214

For more on arithmetic progressions, check: http://www.veritasprep.com/blog/2012/03 ... gressions/

I think I am missing something here: The formula provided in the question is (n/2)*(2a + n-1)d which is different from the formula Karishma provided which is n/2(2a + (n-1)*d). if you apply the formula on each you get different results

1) Given Formula (n/2)*(2a + n-1)d for even integers between 25 and 63 n=19 d=2 a=26
= (19/2)*(2*26 + 18)2 = (19/2)*(52+18)2 = (19/2)*(70)2 = 19*70 = 1330

2) Karishma's forumla n/2(2a + (n-1)*d) for even integers between 25 and 63 n=19 d=2 a=26
= (19/2)(2(26)+(19-1)2) = 19(26+18) = 19*44 = 836

I am sure my mistake is really obvious but I just can't see it. I would appreciate it if you can correct me where I am wrong.
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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oasis90 wrote:
VeritasPrepKarishma wrote:
jainaba wrote:
for the 2nd part of the problem, omitting the even integers 25-63, why does the gmat explanation assign d=2?
I understand assigning a=26 and n=19 but not following why d=2.
Any assistance would be appreciated.

thanks

The first part of the problem is straight forward:
Sum of numbers from 1 to 100 = 100*101/2 = 50*101 = 5050

In the second part, you need this sum

26 + 28 + 30 + ... + 62

This is an arithmetic progression with common difference of 2 (between consecutive terms, there is a difference of 2).
26 + (26+2) + (26+2*2) + .... (26+2*18)
a, a+d, a+2d,..., a+(n-1)d

So n-1 = 18 and hence, n = 19

Using the formula given, you get
Sum = n/2(2a + (n-1)*d) = 19/2 * [2*26 + (19 -1)*2] = 836

So the required sum = 5050 - 836 = 4214

For more on arithmetic progressions, check: http://www.veritasprep.com/blog/2012/03 ... gressions/

I think I am missing something here: The formula provided in the question is (n/2)*(2a + n-1)d which is different from the formula Karishma provided which is n/2(2a + (n-1)*d). if you apply the formula on each you get different results

1) Given Formula (n/2)*(2a + n-1)d for even integers between 25 and 63 n=19 d=2 a=26
= (19/2)*(2*26 + 18)2 = (19/2)*(52+18)2 = (19/2)*(70)2 = 19*70 = 1330

2) Karishma's forumla n/2(2a + (n-1)*d) for even integers between 25 and 63 n=19 d=2 a=26
= (19/2)(2(26)+(19-1)2) = 19(26+18) = 19*44 = 836

I am sure my mistake is really obvious but I just can't see it. I would appreciate it if you can correct me where I am wrong.

The OP has made a mistake while writing down the formula in the question. The correct formula is what I have provided. For more on AP formulas, check:
http://www.veritasprep.com/blog/2012/03 ... gressions/
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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NoHalfMeasures wrote:
I think it is faster to ignore the given formula all together.

Lets use most commonly known formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms
Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050
Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836
Required sum : 5050 - 836 = 4214
Ans: D

I stared at this question and got super confused by the formula. Thank you for clarifying that you can just ignore that. I could have solved this one!
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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Sum = sum from 1 to 100 - sum from 26 to 62
= 100(100+1)/2 - 19(26 +62)/2
= 5050 - 836
= 4214
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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Could someone tell me how did you know that (2a+(n-1)d) was referring to first plus last one?
And what is d?

Thanks
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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NoHalfMeasures wrote:
The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is give by Sn = n/2(2a + (n-1)*d) = . What is the sum of the integers 1 to 100 inclusive, with the even integers between 25 and 63 omitted.

A. 4345
B. 4302
C. 4258
D. 4214
E. 4170

We can find the sum of the integers 1 to 100 inclusive and subtract from it the sum of the even integers between 25 and 63 (i.e., the even integers from 26 to 62 inclusive).

Using the formula given in the question for the sum of the integers 1 to 100 inclusive, we let a = 1, d = 1, and n = 100; thus, the sum is:

100/2[2(1) + (100 - 1)(1)] = 50(101) = 5050

To calculate the sum of the even integers 26 to 62 inclusive, we let a = 26, d = 2, and n = (62 - 26)/2 + 1 = 19; thus, the sum is:

19/2[26(2) + (19 - 1)(2)] = 19/2(88) = 19(44) = 836

Therefore, the sum of the integers from 1 to 100 inclusive, with the sum of the even integers between 25 and 63 omitted, is:

5050 - 836 = 4214

Note: We don’t need to use the formula given in the problem; we can use the basic formula for the sum of an evenly spaced set of numbers, sum = quantity x average. For example, for the sum of the integers 1 to 100 inclusive, quantity = 100 and average = (1 + 100)/2 = 50.5. Thus, sum = 100 x 50.5 = 5050.

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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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Hi,
I do not understand why we substract 836 from 5050. Could anyone help me?

Thank you,
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Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv  [#permalink]

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NoHalfMeasures wrote:
I think it is faster to ignore the given formula all together.

Lets use most commonly known formula for sum of consecutive/evenly spaced numbers: (average of 1st and last no.s)*no of terms
Sum of all integers 1 to 100 inclusive : (1+100)/2 * 100 = 5050
Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836
Required sum : 5050 - 836 = 4214
Ans: D

Hi Guys could someone please explain this part to me "Sum of even no.s from 25 to 63: (26+62)/2 * 19 = 836"?

Best Re: The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is giv   [#permalink] 11 May 2019, 11:57

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