Find the sum of all natural numbers between 403 and 603 which are divisible by 7
Explanation
Given: 403, 404, 405, 406, . . . . 602
Numbers divisible by 7 are 406, 413, 420, . . . . . 602.
S\(_n\) = \(\frac{n}{2}\)[2a + (n - 1)d]
a = 406, d = 7
T\(_n\) = a + (n - 1)d
602 = 406 + (n - 1)7
602 - 406 = 7n - 7
7n = 602 - 406 + 7 = 203
n = \(\frac{203}{7}\) = 29.
S\(_29\) = \(\frac{29}{2}\)[2(406) + (29 - 1)7]
= \(\frac{29}{2}\)[812+ (28 x 7)]
= \(\frac{29}{2}\)[812+ (196)]
= \(\frac{29}{2}\)[1008] = 14616