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 AP, whose first term is $$10\mathrm{a r}, \mathrm{n}^{\text {th }}$$ term is $$\mathrm{a}_{\mathrm{n}}$$ and the common difference is $$10 \mathrm{ar}^{2}$$, is equal to :

Suppose $$a_{1}, a_{2}, \ldots, a_{n}$$, .. be an arithmetic progression of natural numbers. If the ratio of the sum of first five terms to the sum of first nine terms of the progression is $$5: 17$$ and , $$110 < {a_{15}} < 120$$, then the sum of the first ten terms of the progression is equal to

Consider two G.Ps. 2, 2², 2³, ..... and 4, 4², 4³, .... of 60 and n terms respectively. If the geometric mean of all the 60 + n terms is $${(2)^{{{225} \over 8}}}$$, then $$\sum\limits_{k = 1}^n {k(n - k)} $$ is equal to :

The sum $$\sum\limits_{n = 1}^{21} {{3 \over {(4n - 1)(4n + 3)}}} $$ is equal to