Correct Answer: See explanation

Explanation:

The sum of the odd numbers from 1 to 125inclusive = 1 + 3 + 5 + ………+ 125

It’s an arithmetic progressionWhere, a= 1; d=3-1 =2

Here. n-th term = 125

so, a+(n-1)d = 125=> 1+(n-1)2 =125

=> 2n-2 = 125 -1

=> 2n = 124+2

so, n = 63

so, Sum. $s_{n}=\frac{n}{2}[2a+(n-1)d]$

=$\frac{63}{2}[2\times 1+(63-1)\times 2]$

=$\frac{63}{2}$[(2+124)]

= 3969

Again. sum of the odd numbers form 169 to 209 inclusive = 169+171+173+……………………+209

Here. a = 169

d = 171 – 169 = 2

Here, n-th term = 209

so, a+(n-1)d = 209

=> 169+(n-1)2=209

=> 2n-2 = 209-169

=> 2n = 40+2

so, n = 21

Now, Sum $S_{n}=\frac{n}{2}[2a+(n-1)d]$

= $\frac{21}{2}[2\times 169+(21-1)\times 2]$

=3969 [proven]