Let $S_1 = \{100, 105, 110, 115, ... \}$ and $S_2 = \{100, 95, 90, 85, ... \}$ 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 __________.

Let the Arithmetic Progression, $S_1 = \{100, 105, 110, 115, ...\}$

• first term $(a_1) = 100$
  • common difference $(d_1) = 5$

Let the Arithmetic Progression, $S_2 = \{100, 95, 90, 85, ...\}$

• first term $(b_1) = 100$
  • common difference $(d_2) = -5$

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For an arithmetic progression with first term $a$ and common difference $d$, the $k$-th term is:

$T_k = a + (k-1)d$

For $S_1$: $a_k = 100 + (k-1)(5) = \boxed{95 + 5k}$

For $S_2$: $b_k = 100 + (k-1)(-5) = \boxed{105 - 5k}$

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Now let's find the product $a_k b_k$:

$a_k b_k = (95 + 5k)(105 - 5k)$

$a_k b_k = 9,975 + 50k - 25k^2$

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Calculating the sum:

$\Rightarrow \sum_{k=1}^{20} a_k b_k$

$\Rightarrow \sum_{k=1}^{20} (9,975 + 50k - 25k^2)$

This splits into three parts:

$\Rightarrow 20 \times 9,975 + 50\sum_{k=1}^{20} k - 25\sum_{k=1}^{20} k^2$

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Part 1: $20 \times 9,975 = 199,500$

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Part 2: Using $\sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$

$50\sum_{k=1}^{20} k = 50 \times \dfrac{20 \times 21}{2} = 10,500$

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Part 3: Using $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$

$25\sum_{k=1}^{20} k^2 = 25 \times \dfrac{20 \times 21 \times 41}{6} = 71,750$

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Combining all parts:

$\sum_{k=1}^{20} a_k b_k = 199,500 + 10,500 - 71,750 = 138,250$