Find the sum of all 2-digit natural numbers
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10,11,12,.....99 are two digit natural numbers are in arithmetic progression
First term = a =10,
Common difference = d =1
Last term= l = 99
a +(n-1)d =99
10 +n-1 =99
9+n =99
n =99-9
n=90
Therefore number of two digit numbers are 90
Sum of 90 terms = n(a+l)/2
= 90(10+99)/2
=90×109/2
=9810/2
=4905
First term = a =10,
Common difference = d =1
Last term= l = 99
a +(n-1)d =99
10 +n-1 =99
9+n =99
n =99-9
n=90
Therefore number of two digit numbers are 90
Sum of 90 terms = n(a+l)/2
= 90(10+99)/2
=90×109/2
=9810/2
=4905
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