Find the sum of all the three digit numbers which are completely divisible by 5.
Answers
Answered by
0
Answer:
1st 3 digit no = 100,
Last no = 995
Total numbers (let 'n') in btw are…
Last no. = first no + (n - 1) d. {Equation…}
995 = 100 + (n - 1)5
After calculating...we have value of ,'n'.
n = 178.
Sum of these 'n' numbers….(let 'S').
S =( n/2)[ first no + last no ].
S = (178/2)[100 + 995 ]
S= (89)[1095]
S= 97,455.
Therefore, the sum of 3 digit numbers divisible by 5 is 97,455.
____________________
Hope it helps you!!
:)
Similar questions