VJesus12 wrote:The sum S of the arithmetic sequence $$a,\ a+d,\ a+2d,\ \dots\ ,\ a+(n-1)\cdot d$$ is give by $$Sn=\frac{n}{2}(2a+(n-1)\cdot d\ ).$$ 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

The OA is D.

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