S = 2 1 n ( a + l )
a = 11 l = 99 n = 9
S = 2 1 9 ( 11 + 99 ) S = 2 1 9 ( 110 ) S = 2 990 S = 495
11+22+33+44+55+66+77+88+99 33+77+121+165+99 110+286+99 396+99 495
The sum of the two-digit multiples of 11, which include 11, 22, 33, 44, 55, 66, 77, 88, and 99, is 495. This was calculated using the formula for the sum of an arithmetic series. There are a total of 9 terms in this sum.
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