Question

find the sum of all two digit odd positive number

Answer 1

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Answer 2

We know that Sum of the n terms = (n/2)[2a+(n -1)d]    -------------- (1) 

                                             tn = a + (n -1)d

                                            99 = 11 + (n-1)2

                                            99 - 11 = (n-1)2

                                           88/2 = (n-1)

                                            n = 45.

Substitute n in (1), we get

Sum of 45 terms = (45/2) * (2 * 11 + (45 - 1) * 2)

                             = 45 * 55

                            = 2475.

Hope this helps!