1.

The sum of all multiples of 7 between 0 and 500 is

Answer»

A (first term) = 7
l (last term) = 497
d (common difference) = 7

l = a + (n-1) × d
497 = 7 + (n-1) × 7
497-7 = (n-1) ×7
490 = (n-1)×7
490/7 = (n-1)
70 = n-1
n = 70+1
n (no. of terms) = 71

Now sum of multiple =
sn \: = \frac{n}{2} (a + l) \\ \\ sn = \frac{71}{2} (7 + 497) \\ \\ sn = \frac{71}{2} \times (504) \\ \\sn = 71 \times 252 \\ sn = 17892
Therefore, Sum of all multiple = 17892.

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