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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
2 5 2
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2015-05-08T15:48:23+05:30

This Is a Certified Answer

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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

3 5 3
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..:-)