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

QA = 820

a

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.

(40+1)/2((40-1)+1) = 41/2(40) = 820
Quantity A is greater

Posted from my mobile device

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?

\(1+2+3...........+n=\frac{n(n+1)}{2}\)

\(\frac{40(40+1)}{2}=20*41=820\)

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