The sum S of the arithmetic sequence a, a+d,....

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VJesus12: Legendary Member; Posts: 2276; Joined: Sat Oct 14, 2017 6:10 amFollowed by: 3 members

by VJesus12 » Mon Mar 19, 2018 4:09 am

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.

I don't know how to use the given formula. May someone gives me some help? Please.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Mar 19, 2018 4:24 am

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

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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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pmFollowed by: 5 members; Thanked: 6 times

by Vincen » Mon Mar 19, 2018 4:53 am

I would solve it as follows:

Numbers from 1 to 100 are:

1, 2, 3, 4, . . . , 98, 99, 100
1, 1+1, 1+2*1 + 1+3*1 , . . . . , 1+97*1 + 1+98*1, 1+ 99*1.

Hence, a=d=1 and n-1=99, therefore n=100. Now, using the given formula we get: $$S_{100}=\frac{100}{2}(2\cdot1+(100-1)\cdot1\ )\ =\ 50\left(101\right)=5050.$$ Now, the even numbers between 25 and 63 are:

26, 28, 30, . . . . . , 60, 62

26, 26+1*1, 26+2*2, 26+3*2, . . . . . . , 26+17*2, 26+18*2.

Hence a=26, d=2 and n-1=18, therefore n=19. Using the formula we get: $$S=\frac{19}{2}(2\cdot26+(19-1)\cdot2\ )\ =\ \frac{19}{2}\left(88\right)=836.$$ Now, since we have to omitt the even numbers, we have to subtrack 836 from 5050. That is to say, 5050-836=4214.

Thus, the correct answer is the option D.

I hope it helps.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8151; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

by Scott@TargetTestPrep » Fri May 24, 2019 3:12 pm

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

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