Question

what is the sum of all natural number between 70 and 410 which are divisible by 5?
a 16560
b 16561
c 16559
d 16563​

Answer 1

Answer:

First of all you'll have to find out the numbers which are divisible by 6;

so here it is;

252, 258, 264, 270, 276, 282, 288, 294, 300, 306, 312, 318, 324, 330, 336, 342, 348, 354, 360, 366, 372, 378, 384, 390, 396, 402, 408, 414, 420, 426, 432, 438, 444, 450, 456, 462, 468, 474, 480, 486, 492, 498

First term, a = 252

Last term, l = 498

Common difference, d = 6

Let the number of all results take as Ln

Ln= l = a + (n –1) d

⇒ 498 = 252 + (n –1) 6

⇒ 6 (n –1) = 246

⇒ n –1 = 41

⇒ n = 42

Note: If the number of results are less in numbers and if you count the number of results easily that you got then you don't have to follow above equation. you can just simply count and consider the number of results as “n"

so sum of all digits of those numbers;

S = n/2 (a+l)

= 42/2 (252+498)

= 21 (750)

S = 15750

Thus, the required sum is 15750.