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Re: Sum of the integers from 1 to 40 inclusive [#permalink]
1
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]
QA = 820
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Re: Which is greater: Sum of the integers from 1 to 40 inclusive [#permalink]
a
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Sum of the integers from 1 to 40 [#permalink]
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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]
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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]
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]
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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]
1
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]
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