find the sum of 3 digit natural numbers which are divisible by 7
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1
Answer:
70366
Step-by-step explanation:
a=105 tn=994 d=7
tn=a+(n-1)d
994=105+(n-1)7
994-105=n-1×7
889/7=n-1
127+1=n
n=128
Sn=n/2 (a+tn)
s128=128/2 (105+994)
=64 (1099)
=70366
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