For any positive integer n, the sum of the first n positive integers : Problem Solving (PS)

shrouded1

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

22 Oct 2010, 16:03

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metallicafan wrote:

For any positive integer n, the sum of the first n positive integers equals \((n*(n+1))/2\). What is the sum of all the even integers between 99 and 301?
(A) 10,100
(B) 20,200
(C) 22,650
(D) 40,200
(E) 45,150

Sum = 100+102+...+300
= 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms]
= 100*101 + 2*(1+2+..+100) [Taking 2 common]
= 100*101 + 2*100*101/2 [Using formula given]
= 2*100*101
= 20200

Answer is (b)

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

19 Aug 2020, 07:53

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Expert Reply

Sum of all the even integers between 99 and 301:

The first even integer will be 100 and the last will be 300.

The common difference will be 2.

Number of terms: 300 = 100 + (n-1)2

=> 200 = 2n-2

=> 202 = 2n

=> n = 101

Sum: \(\frac{n}{2}\)[first term + last term]

=> \(\frac{101}{2}\)[100+300]

=> \(\frac{101}{2}\) * 400

=> 101 *200

=> 20200

Answer B

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

11 Jul 2010, 06:36

thanks!! it makes sense now...what about if the problem was changed to ask for the sum of odd integers? can we amend the formula to get it? thanks!!

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

07 Dec 2010, 07:30

can someone explain me why do we subtract 49 and no 50 ?

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

05 Feb 2011, 18:45

If there are 100 even integers in the sequence why do you +1 and multiply the average (200) by 101?

B

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

21 May 2023, 08:32

To find the sum of all the even integers between 99 and 301, we first need to determine the number of even integers in that range.

The largest even integer within the given range is 300, and the smallest even integer is 100. The set of even integers between 99 and 301 can be expressed as {100, 102, 104, ..., 298, 300}.To find the number of terms in this arithmetic sequence, we can use the formula for the nth term of an arithmetic sequence:

an = a1 + (n - 1)d

In this case, a1 = 100 (the first term), d = 2 (the common difference), and we need to find n.

300 = 100 + (n - 1)2
200 = 2n - 2
202 = 2n
n = 101

Therefore, there are 101 even integers between 99 and 301.

Now, we can calculate the sum using the formula for the sum of an arithmetic series:

Sn = (n/2)(a1 + an)
Sn = (101/2)(100 + 300)
Sn = (101/2)(400)
Sn = 101 * 200
Sn = 20,200

The sum of all the even integers between 99 and 301 is 20,200.

Therefore, the answer is option B.

B

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

30 Jun 2023, 06:35

Bunuel wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Approach #1:

Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: https://gmatclub.com/forum/math-number- ... 88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: https://gmatclub.com/forum/totally-basi ... ml#p730075);

The sum = 200*101= 20,200.

Answer: B.

Approach #2:

Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Answer: B.

Hope it helps.

In approach no. 1 - what would have been the difference if they wouldnt have mentiond 'even' numbers?

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

30 Jun 2023, 08:16

Expert Reply

Rajeet123 wrote:

Bunuel wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Approach #1:

Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: https://gmatclub.com/forum/math-number- ... 88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: https://gmatclub.com/forum/totally-basi ... ml#p730075);

The sum = 200*101= 20,200.

Answer: B.

Approach #2:

Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Answer: B.

Hope it helps.

In approach no. 1 - what would have been the difference if they wouldnt have mentiond 'even' numbers?

If we were asked to find the sum of the integers between 99 and 301, inclusive, we can utilize the fact that the sum of the elements in any evenly spaced set is given by: \(Sum=\frac{a_1+a_n}{2}*n\), the mean multiplied by the number of terms. Therefore, the answer would be (99 + 301)/2*203 = 40,600.

L

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

30 Jun 2023, 09:34

Given: For any positive integer n, the sum of the first n positive integers equals n(n+1)/2.
Asked: What is the sum of all the even integers between 99 and 301?

Sum of all even integers between 99 and 301 = 100 + 102 + .... + 300 = (2 + 4 + .... + 300) - (2 + 4 + ... + 98) =
= 2 (1+ 2+ .... + 150) - 2 (1+2+... + 49) = 150*151 - 49*50 = 50 (453 - 49) = 50*404 = 20200

IMO B

S

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Re: For any positive integer n, the sum of the first n positive integers [#permalink]

04 Apr 2024, 07:26

My method was similar to approach #1 here, please let me know if this is wrong

No. of terms = 300-100 + 1 = 201 terms

Sum of 201 terms in this sequence = (201(202)) / 2 = 20200

Bunuel wrote:

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all the even integers between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

Approach #1:

Even integer between 99 and 301 represent evenly spaced set (aka arithmetic progression): 100, 102, 104, ..., 300. Now, the sum of the elements in any evenly spaced set is the mean (average) multiplied by the number of terms. (Check Number Theory chapter of Math Book for more: https://gmatclub.com/forum/math-number- ... 88376.html)

Average of the set: (largest+smallest)/2=(300+100)/2=200;
# of terms: (largest-smallest)/2+1=(300-100)/2+1=101 (check this: https://gmatclub.com/forum/totally-basi ... ml#p730075);

The sum = 200*101= 20,200.

Answer: B.

Approach #2:

Using the formula of the sum of the first n positive integers: n(n+1)/2.

100+102+...+300=2(50+51+..+150). Now, the sum of the integers from 50 to 150, inclusive equals to the sum of the integers from 1 to 150, inclusive minus the sum of the integers from 1 to 49, inclusive. 2(50+51+..+150)=2*(150(150+1)/2-49(49+1)/2)=20,200.

Answer: B.

Hope it helps.