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Which is greater: Sum of the integers from 1 to 40 inclusive

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Bunuel
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Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink] New post   Updated on: 30 Dec 2022, 08:37
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Quantity A
Quantity B
Sum of the integers from 1 to 40 inclusive
800


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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A

Originally posted by Bunuel on 04 Dec 2017, 04:54.
Last edited by Carcass on 30 Dec 2022, 08:37, edited 1 time in total.
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Re: Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink] New post   04 Dec 2017, 22:09
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Bunuel wrote:
Quantity A
Quantity B
Sum of the integers from 1 to 40 inclusive
800


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.



Here to know the sum of n numbers both inclusive is given by = \(\frac{n(n+1)}{2}\)

So here n = 40

Therefore sum =\(\frac{(40 * 41)}{2}\) = 820.

Hence option A.
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Sum of the integers from 1 to 40 inclusive [#permalink] New post   23 Nov 2018, 03:59
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Quantity A
Quantity B
Sum of the integers from 1 to 40 inclusive
800


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
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Re: Sum of the integers from 1 to 40 inclusive [#permalink] New post   23 Nov 2018, 10:15
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Carcass wrote:
Quantity A
Quantity B
Sum of the integers from 1 to 40 inclusive
800


A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given


from the equation n(n+1)/2 = 40 * 41/2 = 41 * 20 > 40 * 20

Or since this is a consecutive integer form we can find the average as (last + first)/2

And the number of terms as (last - first) = 40 - 1 = 39 + 1 = 40

Answer choice A
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Re: Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink] New post   17 Sep 2019, 02:08
QA = 820
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Re: Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink] New post   30 Nov 2021, 11:14
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Sum of the integers from 1 to 40 [#permalink] New post   21 Oct 2022, 04:17
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Quantity A
Quantity B
Sum of the integers from 1 to 40
800



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


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Re: Sum of the integers from 1 to 40 [#permalink] New post   21 Oct 2022, 12:23
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(40+1)/2((40-1)+1) = 41/2(40) = 820
Quantity A is greater

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Re: Sum of the integers from 1 to 40 [#permalink] New post   28 Oct 2022, 16:50
Quick question so I have the understanding:
sum of arithmetic sequence = (n/2)(a+L)
where n = number in the sequence, a = first number, L = last number

would n = 40?
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Re: Sum of the integers from 1 to 40 [#permalink] New post   28 Oct 2022, 23:37
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\(1+2+3...........+n=\frac{n(n+1)}{2}\)

\(\frac{40(40+1)}{2}=20*41=820\)
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Re: Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink] New post   09 Jan 2025, 06:02
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Quantity A = n*(n+1)/2 =40*41/2 =820
Quantity B = 800
Quant A > Quant B
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Re: Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink]
09 Jan 2025, 06:02
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