Find the sum of all natural numbers between 403 and 603 which are divisible by 7

Assessment: WAEC SSCE - Further Mathematics - 2024 (Essay) Subject: Further Mathematics

Question 1 Report

Find the sum of all natural numbers between 403 and 603 which are divisible by 7

Answer Details

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

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