Let the sum of an infinite G.P., whose first term is $$a$$ and the common ratio is $$r$$, be 5. Let the sum of its first five terms be $$\frac{98}{25}$$. Then the sum of the first 21 terms of an A.P., whose first term is $$10ar$$, $$n^{th}$$ term is $$a_n$$ and the common difference is $$10ar^2$$, is equal to

We first determine the first term $$a$$ and common ratio $$r$$ of the geometric progression using the given sums. Since the sum to infinity is 5, we have $$\dfrac{a}{1-r} = 5$$, and the sum of the first 5 terms is $$\dfrac{a(1-r^5)}{1-r} = \dfrac{98}{25}$$. Dividing the second equation by the first yields

$$1 - r^5 = \frac{98}{125}$$

$$r^5 = 1 - \frac{98}{125} = \frac{27}{125} = \left(\frac{3}{5}\right)^5$$

so $$r = \dfrac{3}{5}$$. Substituting into $$\dfrac{a}{1-3/5} = 5$$ gives $$\dfrac{a}{2/5} = 5 \implies a = 2$$.

The arithmetic progression is formed by taking terms as $$10ar$$ and $$10ar^2$$. Its first term is $$10ar = 10 \times 2 \times \dfrac{3}{5} = 12$$ and its common difference is $$10ar^2 = 10 \times 2 \times \dfrac{9}{25} = \dfrac{36}{5}$$.

For an A.P. with an odd number of terms $$n = 2k+1$$, the sum is $$n$$ times the middle term. With 21 terms, the middle (11th) term is

$$a_{11} = 12 + 10 \times \frac{36}{5} = 12 + 72 = 84$$

Therefore, $$S_{21} = 21 \times a_{11} = \boxed{21a_{11}}$$.

The answer is Option A.