Question

Find the sum of all 2 digit positive integers​

Answer 1

Answer:

The sum of all 2 digit positive integer is

10+11+12+..........+99

Which are in ap

With a=10

d=1

n=89

Sn=n/2{2a+(n-1)d}

=89/2{2x10+(89-1)1}

=44.5(20+88)

=44.5x118

=5251

Answer 2

$\huge\underline{\mathbb\red{➣s}\green{o}\mathbb\blue{l}\purple{u}\mathbb\orange{ti}\pink{on}}\:$,

✰ $\textit{✰two digit postive interges,}$

✰ $\textit{10,11,12.......99}$

$\textit{❥first finding(n) sum of number}$

➠$\textit{99 = 10+(n-1)×1}$

➠$\textit{99 = 10+n - 1}$

➠$\textit{99-9 = n}$

➠$\textit{90= n}$,

$\textit{d = 11-10=1}$

$\textit{a = 10}$

$\textit{❥sum of number s}$

➠$\textit{sn= n/2+(a+an}$

➠ $\textit{sn = 90/2×( 10+99)}$

➠ $\textit{sn = 45×109 = 4950answer.}$