Two arithmetic sequences, with a twist

The sum is given by:

$\begin{aligned} S & = \frac {x(18+3(x-1))}2 + \frac {\color{#D61F06}{y}(0+5(\color{#D61F06}{y}-1))}2 & \small \color{#3D99F6}{\text{Using }S = \frac {n(2a+l)}2 = \frac {n(2a+(n-1)d)}2} \\ & = \frac {3x^2 + 15x}2 + \frac {5(\color{#D61F06}{10-x})(\color{#D61F06}{10-x}-1)}2 & \small \color{#D61F06}{\text{Note that } x + y = 10 \implies y = 10-x} \\ & = \frac {3x^2 + 15x}2 + \frac {5x^2 - 95x+450}2 \\ & = 4x^2 - 40x + 225 \\ & = 4(x^2 -10x + 25) + 125 \\ & = 4\color{#3D99F6}{(x-5)^2} + 125 \end{aligned}$

Since $(x-5)^2 \ge 0$ , $S$ is minimum, when $(x-5)^2=0$ or $x=5$ and the minimum value of $S_{min} = \boxed{125}$

We express the sum of the first $x$ terms of $\{a_n\}$ and the first $y$ terms of $\{b_n\}$ in terms of $x$ and $y$ respectively.

We know that the sum of the first $n$ terms of an arithmetic sequence with first term $a_1$ and common difference $d$ is given by

$\frac{(n)(2a_1+(n-1)d)}{2}$

We substitute the givens in the sequences $\{a_n\}$ and $\{b_n\}$ to get the requested sum from the problem to be

$\frac{3}{2} (x)(x+5) + \frac{5}{2} (y)(y-1)$

Doing the substitution $y = 10-x$ and expanding gives us that the requested sum becomes

$\frac{3}{2} (x)(x+5) + \frac{5}{2} (y)(y-1) = \frac{3}{2} (x)(x+5) + \frac{5}{2} (10-x)(9-y) = 4x^2-40x+225$

Which is a quadratic equation which obtains its minimum when $x = -\frac{-40}{2(4)} = 5$

Plugging in $x = 5$ into the formula gives us the minimum value to be $\boxed{125}$ .

P.S. This is another inequality which gives us the minimum when $x=y$ .