Find the sum of all 3 digit number which are divisible by 5
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Answers
Answered by
1
Answer:
a=100
d=5
l=995
l=a+(n-1)d
995=100+5n-5
900=5n
n=180
S=n/2[2a+(n-1)d]
S=180/2[2*100+179*5]
S=90[200+895]
S=90*1095
S=98550
Step-by-step explanation:
Answered by
0
Answer:
98550
Step-by-step explanation:
first; lets list the three digit numbers divisible by 5
100
105
200
205
.
.
.
995
we observe that this series is an A.P. with first term 100 and last term 995 and common difference as 5
threfore we first try to find the total number of terms involved
An=a+(n-1)d
995=100+(n-1)5
895/5=n-1
179+1=n
180=n
now; applying the formula for sum of A.P.
S=n/2(a+l)
S=180/2(100+995)
S=90(1095)
S=98550
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