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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

10 + 11 + ... + 99

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

s = 99 x 100/2 - 45

s = 4950 - 45 = 4905

__ANSWER__### This Is a Certified Answer

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 isSum 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]

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

Sn = 45(109)

Sn = 4905