how many three digit natural no divisible by 5 ?
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since, 100,105,110.....995are the three digit natural divisible by 5 which is an AP
here, a=100, d= 5 , and tn= 995
we have
tn= a+(n-1)d
995=100+(n-1)5
995-100=5n-5
895+5=5n
900=5n
n=180
therefore, there are 180 three digit natural no. which are divisible by 5
here, a=100, d= 5 , and tn= 995
we have
tn= a+(n-1)d
995=100+(n-1)5
995-100=5n-5
895+5=5n
900=5n
n=180
therefore, there are 180 three digit natural no. which are divisible by 5
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