Let $$S_1 = \left\{100, 105, 110, 115, ...\right\}$$ and $$S_2 = \left\{100, 95, 90, 85, ...\right\}$$ be two series in arithmetic progression. If $$a_k$$ and $$b_k$$ are the $$k^{th}$$ terms of
$$S_1$$ and $$S_2$$, respectively, then $$\sum_{k=1}^{20}a_k b_k$$ equals ____________.

Since $$a_k$$ is the $$k^{th}$$term of $$S_1$$

So, $$a_k=100+5\left(k-1\right)=95+5k$$

Similarly, $$b_k=100-5\left(k-1\right)=105-5k$$

So, $$\sum_{k=1}^{20}a_k b_k$$ = $$\sum_{k=1}^{20}\left(105-5k\right)\left(95+5k\right)$$

=$$5\times\ 5\sum_{k=1}^{20}\left(21-k\right)\left(19+k\right)$$

=$$25\sum_{k=1}^{20}\left(399+21k-19k-k^2\right)$$

=$$25\left(\sum_{k=1}^{20}\left(399+2k-k^2\right)\right)$$

=$$25\left(\sum_{k=1}^{20}399+2\sum_{k=1}^{20}k-\sum_{k=1}^{20}k^2\right)$$

Now, we know $$\sum_{k=1}^nk=\dfrac{n\left(n+1\right)}{2}$$ and $$\sum_{k=1}^nk^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}$$ and here $$n=20$$

=$$25\left(\ 399\times\ 20+2\times\ 20\times\ \dfrac{21}{2}-20\times\ 21\times\ \dfrac{41}{6}\right)$$

=$$25\left(399\times\ 20+420-2870\right)$$

=$$25\times\ 5530$$

=$$138250$$