Question

if the 2003rd positive odd number is N then sum of digits of N is

Answer 1

Answer:

N is 4005 and sum of its digit is 9

Step-by-step explanation:

2003rd positive number

$t_{n} = a+ (n-1)*d\\ n = 2003 , a=1 , d= 2\\ t_{2003} = 1+ (2003-1) * 2$

         = 1+2002×2

          = 4005

sum of digit of 4005 is 4+0+0+5 = 9

I hope it will help you!