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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The sum S of the arithmetic sequence a, a+d,....
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For any EVENLY SPACED SET: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
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
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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.
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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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.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.
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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