Question

Find the sum of all 2-digit natural numbers

Answer 1

so all 2 digit natural numbers is 

10 + 11 + ... + 99

so s = n(n+1)/2 - 45

s = 99 x 100/2 - 45 

s = 4950 - 45 = 4905 ANSWER

Answer 2

1st method:

Sum of 'n' numbers is
Sn = n/2 (2a+(n-1)d)
first 2 digit number = 10, difference between any 2 successive numbers = 1, last number= 99, no.of digits from 10 to 99 = 90
Sn = 90/2 [2(10)+(90-1)1]
Sn = 45[20+89]
Sn = 45[109]
Sn = 4905
therefore, sum of all 2 digit numbers = 4905

2nd method:

Sn = n/2 [a+l]
first 2 digit number (a) = 10, last number(l) = 99, no.of digits from 10 to 99 = 90
Sn = 90/2 [10+99]
Sn = 45(109)
Sn = 4905