# Find the sum of all 2-digit natural numbers

2
by NealLalla498

## Answers

2015-05-08T15:44:58+05:30

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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
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2015-05-08T15:48:23+05:30

### This Is a Certified Answer

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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

i am doing second method too
or u can use the modified formula of Sn.. ie Sn= (n/2)(a+l)
yup i did with that.. :)
yeah..:-)