All the two-digit odd positive numbers are –

11,13,15,17,……….,99

The above series of numbers forms an arithmetic progression with

first term(a) = 11 and,

common difference(d) = (n + 1)th term – nth term = 13 – 11 = 2

last term or nth term(a_n) = 99

Let the total no. of terms in above A.P be n.

∴ a_n = a + (n – 1) × d

⇒ 99 = 11 + (n – 1) × 2

⇒ 88 = 2n – 2

⇒ 2n = 90

∴ n = 45

Sum of all the 45 terms of the AP is given by –

S₄₅ =(45/2)(11 + 99)

[∵S_n = (n/2)(a + l) =(n/2)[(2a + (n – 1)d]

=(45/2) × 110

=45 × 55

=2475

Thus, the sum of all two-digit odd positive numbers = 2475.