Sir/Madam Plss solve it

Sum of all integers between 100 and 500 which are divisible by 17

Asked by seeni2005 | 17th Aug, 2020, 12:42: AM

Expert Answer:

The first and last numbers between 100 and 500 which are divisible by 17 are 102 and 493.
The numbers divisible by 17 forms an AP with common difference d=17, first term a = 102 and last term an = 493
an = a + (n-1)d
493 = 102 + (n-1)17
(n-1)17 = 391
n-1 = 23
n = 24
Now, the formula for sum of n terms of an AP can be used to find the required sum
straight S subscript straight n equals straight n over 2 open square brackets 2 straight a plus open parentheses straight n minus 1 close parentheses straight d close square brackets
rightwards double arrow straight S subscript 24 equals 24 over 2 open square brackets 2 cross times 102 plus open parentheses 24 minus 1 close parentheses 17 close square brackets
equals 12 open square brackets 204 plus 23 cross times 17 close square brackets
equals 12 open parentheses 204 plus 391 close parentheses
equals 12 open parentheses 595 close parentheses
equals 7140

Answered by Renu Varma | 17th Aug, 2020, 12:19: PM