Consider an $$A.P:a_{1},a_{2},...a_{n};a_{1} > 0$$. If $$a_{2}-a_{1}=\frac{-3}{4},a_{n}=\frac{1}{4}a_{1}$$, and $$\sum_{i=1}^{n}a_{i}=\frac{525}{2}$$, then $$\sum_{i=1}^{17}a_{i}$$ is equal to

We need to find $$\sum_{i=1}^{17} a_i$$ for an A.P. with $$a_1 > 0$$, $$a_2 - a_1 = -\frac{3}{4}$$, $$a_n = \frac{1}{4}a_1$$, and $$\sum_{i=1}^n a_i = \frac{525}{2}$$.

First, the common difference is $$d = a_2 - a_1 = -\frac{3}{4}$$, and since $$a_n = a_1 + (n-1)d = \frac{1}{4}a_1$$ one has $$a_1 + (n-1)\left(-\frac{3}{4}\right) = \frac{a_1}{4}$$, which simplifies to $$\frac{3a_1}{4} = \frac{3(n-1)}{4}$$ hence $$a_1 = n-1$$.

Next, the sum formula $$S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}\left(a_1 + \frac{a_1}{4}\right) = \frac{5na_1}{8}$$ equals $$\frac{525}{2}$$, so $$na_1 = \frac{525 \times 8}{2 \times 5} = 420$$. Since $$a_1 = n-1$$, we solve $$n(n-1) = 420$$ giving $$n^2 - n - 420 = 0$$ and $$n = \frac{1+\sqrt{1+1680}}{2} = \frac{1+41}{2} = 21$$, thus $$a_1 = 20$$.

Then to compute the required sum, use $$S_{17} = \frac{17}{2}(2a_1 + 16d) = \frac{17}{2}\left(40 + 16 \times \left(-\frac{3}{4}\right)\right) = \frac{17}{2}(40 - 12) = \frac{17 \times 28}{2} = 238$$.

Therefore, $$\sum_{i=1}^{17} a_i = 238$$, which matches Option C.