find how many three digit natural numbers are divisible by 5
Answers
Answered by
122
100,105........995
a=100
d=105-100=5
an=995
an=a+(n-1)d
995=100+(n-1)5
995-100=(n-1)5
895=(n-1)5
895/5=n-1
179=n-1
n=179+1
n=180
hope this helps you
a=100
d=105-100=5
an=995
an=a+(n-1)d
995=100+(n-1)5
995-100=(n-1)5
895=(n-1)5
895/5=n-1
179=n-1
n=179+1
n=180
hope this helps you
Answered by
80
yup!
its me Nishant !!
SOLUTION⏩
SOLVED BY 【 A•P】
A•P: 100,105,110........955
First term=100
last term=995
common difference=5
let the total no. of three digit term divisible by 5 be n. then,
tn=>a +(n-1)d=995
>>=> 100+(n-1)5=995
>>=>(n-1)5=895
>>=>n-1=179
>>=>n=180
➡HENCE,TOTAL NO. OF THREE DIGIT NUMBER DIVISIBLE BY 5 IS 180
HOPE IT HELPS U✌
its me Nishant !!
SOLUTION⏩
SOLVED BY 【 A•P】
A•P: 100,105,110........955
First term=100
last term=995
common difference=5
let the total no. of three digit term divisible by 5 be n. then,
tn=>a +(n-1)d=995
>>=> 100+(n-1)5=995
>>=>(n-1)5=895
>>=>n-1=179
>>=>n=180
➡HENCE,TOTAL NO. OF THREE DIGIT NUMBER DIVISIBLE BY 5 IS 180
HOPE IT HELPS U✌
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