Bunuel wrote:
What is the sum of the first 50 multiples of a positive integer K?
(A) 1,225K
(B) 1,275K
(C) 2,450K
(D) 2,550K
(E) 1,326K
Zero is a multiple of every integer.
The first 50 multiples of\(K\) are
\(0K, 1K, 2K, 3K, . . . . 47K, 48K, 49K\)
Factor K out (we're interested in the arithmetic progression)
Sum of first 50 multiples of\(K\):
\(K(0 + 1 + 2 + 3 . . . + 47 + 48 + 49)\)
Sum of evenly spaced integers=
(Average)*(# of terms)
Average = \(\frac{FirstTerm+LastTerm}{2}\)
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