Question

Find the sum of all two digit positive number

Answer 1

33+11=44;11+12=23............

Answer 2

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$two \: digit \: number \: are \: 10 ,\: 11 ,\: ,12.........,99 \\ these \: number \: forms \: an \: ap \: with \: a = 10 \: \: d = 1 \: and \: l = 99 \\ let \: the \: number \: o f\: these \: number \: be \: n \: \: \: \: \: \: then \\ 99 = 10 + (n - 1) \times 1 \implies89 = n - 1 \implies90 \\ \therefore \: sum \: of \: these \: numbers \: = \frac{90}{2} (10 + 99) = 45 \times 109 = 4905$