(A) : The sum 1 + 3 + 5 + .... + 21 = 100
(R) : The sum of first $$n$$ odd positive integers is $$n^2$$

We know that the nth odd number is 2n−1 and the nth even number is 2n

$$Now, let's assume the sum of first n odd numbers to be S i.e. S=1+3+5+…+(2n−1)$$

Now let us add 1 n times to the right side,

$$S+n=(1+1)+(3+1)+(5+1)+…+(2n−1+1)$$

$$or,S+n=2+4+6+…+2n$$

Now adding these 2 equations, we get,

$$2S+n=1+2+3+…+(2n−1)+2n$$

The RHS is the sum of first 2n natural numbers which is as below,

$$2S+n=2n(2n+1)/2$$

$$or,2S+n=2n^2+n$$

$$or,2S=2n^2$$

$$or,S=n^2$$

$$So the sum of first n odd integers is n^2.$$

So, Condition R is true.

But Condition A is not true.

So, Option D is correct choice.