Correct Answer  Option 3 : 247500
Concept:
Sum of n Terms of AP is given by:
\(\rm s_{n}= \frac{n}{2}[2a + (n  1)d]\)
Where, a = First term, d = Common difference and n = number of terms
an = n^{th} term = a + (n  1) d
Calculation:
We can use the concept of arithmetic progression and find the sum of n^{th} term.
Three digit odd numbers are 101, 103, ..., 999
a (first term) = 101, n^{th} term (last term) = 999, d = 2
As we know, an = nth term = a + (n  1) d
⇒ 999 = 101 + (n  1) × 2
⇒ 898 = (n  1) × 2
∴ n = 450
The formula for finding the sum of n^{th} terms of A.P. is:
\(\rm s_{n}= \frac{n}{2}[2a + (n  1)d]\)
\(\rm s_{450}= \frac{450}{2}[2 × 101 + (450  1)2]\)
\(\rm s_{450}= 225 ×[202 + 449×2]\)
\(\rm s_{450}= 225 ×[202 + 898]\)
\(\rm s_{450}= 225 × 1100\)
\(\rm s_{450}= 247500\)
The list of formulas is given in a tabular form used in AP. These formulas are useful to solve problems based on the series and sequence concept.
General Form of AP 
a, a + d, a + 2d, a + 3d, . . . 
The nth term of AP 
an = a + (n – 1) × d

Sum of n terms in AP 
S = n/2[2a + (n − 1) × d] 
Sum of all terms in a finite AP with the last term as ‘l’ 
n/2(a + l) 