The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-

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The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is given 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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by GMATGuruNY » Thu May 10, 2018 4:04 pm
alanforde800Maximus wrote:The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is given 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
For any EVENLY SPACED SET:
Count = (biggest - smallest)/(increment) + 1.
Average = (biggest + smallest)/2.
Sum = (count)(average).
The INCREMENT is the difference between successive values.

Integers between 1 and 100, inclusive:
Here, the integers are CONSECUTIVE, so the increment = 1.
Count = (100-1)/1 + 1 = 100.
Average = (100 + 1)/2 = 101/2.
Sum = (100)(101/2) = 5050.

Even integers between 26 and 62, inclusive:
Here, the integers are EVEN, so the increment = 2.
Count = (62-26)/2 + 1 = 19.
Average = (62+26)/2= 44.
Sum = (19)(44) = (20-1)(44) = 880-44 = 836.

Subtracting the second sum from the first, we get:
5050 - 836 = 4214.

The correct answer is D.
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by alanforde800Maximus » Thu May 10, 2018 5:14 pm
GMATGuruNY wrote:
alanforde800Maximus wrote:The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is given 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
For any EVENLY SPACED SET:
Count = (biggest - smallest)/(increment) + 1.
Average = (biggest + smallest)/2.
Sum = (count)(average).
The INCREMENT is the difference between successive values.

Integers between 1 and 100, inclusive:
Here, the integers are CONSECUTIVE, so the increment = 1.
Count = (100-1)/1 + 1 = 100.
Average = (100 + 1)/2 = 101/2.
Sum = (100)(101/2) = 5050.

Even integers between 26 and 62, inclusive:
Here, the integers are EVEN, so the increment = 2.
Count = (62-26)/2 + 1 = 19.
Average = (62+26)/2= 44.
Sum = (19)(44) = (20-1)(44) = 880-44 = 836.

Subtracting the second sum from the first, we get:
5050 - 836 = 4214.

The correct answer is D.
Thanks Mitch.

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by Scott@TargetTestPrep » Mon May 14, 2018 4:05 pm
alanforde800Maximus wrote:The sum S of the arithmetic sequence a, a+d, a+2d,..., a+(n-1)d is given 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 see that the required sum is the sum of the integers 1 to 100 inclusive minus the sum of the even integers from 26 to 62 inclusive.

Notice that Sn = n/2(2a + (n-1)*d) = n/2(a + [a + (n-1)d]). That is, the sum of the first n terms of an arithmetic sequence is n/2 times the sum of the first and last terms of the sequence.

Thus, for the sum of the integers 1 to 100 inclusive, there are n = 100 terms, the first term is 1 and the last term is 100. Therefore, the sum = 100/2 x (1 + 100) = 50 x 101 = 5050.

Similarly, for the sum of the even integers 26 to 62 inclusive, there are n = (62 - 26)/2 + 1 = 19 terms, the first term is 26 and the last term is 62. Therefore, the sum = 19/2 x (26 + 62) = 19/2 x 88 = 19 x 44 = 836.

Thus, the required sum is 5050 - 836 = 4214.

Answer: D

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