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the sum s of the arithmetic sequence a, a+d, a+2d........+ a

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the sum s of the arithmetic sequence a, a+d, a+2d........+ a [#permalink] New post 23 Feb 2018, 03:51
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the sum s of the arithmetic sequence a, a+d, a+2d........+ a + (n-1)d is given by \(S = \frac{n}{2} * [2a + (n-1)d]\). What is the sum of the integers from 1 to 100, inclusive with the even integers between 25 and 63 omitted.

A) 4346
B) 4302
C) 4258
D) 4214
E) 4170
[Reveal] Spoiler: OA
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Re: the sum s of the arithmetic sequence a, a+d, a+2d........+ a [#permalink] New post 23 Feb 2018, 08:13
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To find the sum of integers from 1 to 100 and omitting even integers between 25 to 63, we first find the sum of all integers from 1 to 100:
S=n*(n+1)/2

S=100*101/2
S=5050

Then we find the sum of all even integers between 25 and 63 using the given formula for an AP:
X=(n/2)*[2a+(n-1)d]; (we know a=26, d=2) ---------------1

to find n:
We know that the last even number to be added=62
62=26+(n-1)*2 => 38=2*n => n=19

Putting these values in equation 1:
X=(19/2)*[2*26+(18)*2]
X=836

Subtracting this number from the entire sum:

answer=5050-836=4214
answer=D
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Re: the sum s of the arithmetic sequence a, a+d, a+2d........+ a [#permalink] New post 14 May 2018, 05:13
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amorphous wrote:
the sum s of the arithmetic sequence a, a+d, a+2d........+ a + (n-1)d is given by \(S = \frac{n}{2} * [2a + (n-1)d]\). What is the sum of the integers from 1 to 100, inclusive with the even integers between 25 and 63 omitted.

A) 4346
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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Re: the sum s of the arithmetic sequence a, a+d, a+2d........+ a   [#permalink] 14 May 2018, 05:13
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