Question

Find the sum of all multiple of 7 between 1 and 500

Answer 1

388,521

Explanation:

The multiples of 7 between 1 and 500 are:

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497

Add all of these numbers.

7 + 14 + 21 + 28 + 35 + 42 + 49 + 56 + 63 + 70 + 77 + 84 + 91 + 98 + 105 + 112 + 119 + 126 + 133 + 140 + 147 + 154 + 161 + 168 + 175 + 182 + 189 + 196 + 203 + 210 + 217 + 224 + 231 + 238 + 245 + 252 + 259 + 266 + 273 + 280 + 287 + 294 + 301 + 308 + 315 + 322 + 329 +336 + 343 + 350 + 357 + 364 + 371 378 + 385 + 392 + 399 + 406 + 413 + 420 + 427 + 434 + 441 + 448 + 455 + 462 + 469 + 476 + 483 + 490 + 497

=388,521

Answer 2

Answer: 17892

Solution:

The multiples of 7 between 0 to 500 are:

7x1 = 7

7x2 = 14

7x3 = 21

7x4 = 28

7x5 = 35

………….

7x70 = 490

7x71 = 497

Therefore total number of multiples of 7 between 0 and 500 are 71. If we denote by S71 the sum of all these multiples,

S71 = 7+14+21+28+35+…………+490+497

= 7(1+2+3+4+………………….+70+71)

The series within parentheses is the sum of the first 71 natural numbers and the sum is given by the formula n(n+1)/2, where n=total number of terms=71 . Substituting for n,

S71 = (7x71x72)/2 = 7x71x36 = 17892 (Answer

Explanation: