Upstairs Downstairs

It is given that $\begin{cases} a_1+a_2+a_3 = 126 & ...(1) \\ a_1+b_1 = 85 & ...(2) \\ a_2+b_2 = 76 & ...(3) \\ a_3+b_3 = 84 & ...(4) \end{cases}$

$\begin{aligned} (1): \quad a_1+a_2+a_3 & = 126 \quad \quad \small \color{#3D99F6}{\text{For AP: } a_1+a_3 = 2 a_2} \\ \Rightarrow 3 a_2 & = 126 \\ a_2 & = 42 \\ (3): \quad a_2+b_2 & = 76 \\ \Rightarrow b_2 & = 34 \\ (2)+(3)+(4): \quad a_1+a_2+a_3 + b_1+b_2+b_3 & = 85+76+84 \\ 126 + b_1+b_2+b_3 & = 245 \\ b_1+b_2+b_3 & = 119 \quad \quad \small \color{#3D99F6}{\text{Let the common ratio of the GP be }r} \\ b_2 \left( \frac{1}{r} + 1 + r\right) & = 119 \\ 34 \left( \frac{1}{r} + 1 + r\right) & = 119 \\ 2r^2 - 5r + 2 & = 0 \\ \Rightarrow (2r-1) (r-2) & = 0 \end{aligned}$

For $a_1<a_2<a_3$ and comparing $a_1+b_1 = 85$ and $a_3+b_3 = 84$ , we note that $b_1 > b_2 > b_3$ . So, $r=\frac{1}{2}$ and:

$\begin{cases} b_1 = \dfrac{b_2}{r} = 68 \\ b_3 = r b_2 = 17 \end{cases} \quad \Rightarrow \begin{cases} a_1+b_1 = 85 & \Rightarrow a_1 = 17 \\ a_3+b_3 = 84 & \Rightarrow a_3 = 67 \end{cases} \quad \Rightarrow a_3-a_1+b_1-b_3 = \boxed{101}$

Let $a_{1} = a - d, a_{2} = a$ and $a_{3} = a + d$ . Then from condition (i) we have that $3a = 126 \Longrightarrow a = 42.$

Now let $b_{1} = b, b_{2} = br$ and $b_{3} = br^{2}$ . Then from condition (ii) we have that

$a - d + b = 85 \Longrightarrow b - d = 85 - 42 = 43$ ,

and from condition (iv) we have that $a + d + br^{2} = 84 \Longrightarrow br^{2} + d = 42.$

Adding these two results gives us that $b + br^{2} = 85 \Longrightarrow b = \dfrac{85}{1 + r^{2}}$ .

Next, condition (iii) gives us that

$a_{2} + b_{2} = 76 \Longrightarrow a + br = 42 + br = 76 \Longrightarrow b = \dfrac{34}{r}.$

Equating these expressions for $b$ yields that

$\dfrac{34}{r} = \dfrac{85}{1 + r^{2}} \Longrightarrow 34 + 34r^{2} = 85r \Longrightarrow 2r^{2} - 5r + 2 = (2r - 1)(r - 2) = 0$ ,

and so either $r = \dfrac{1}{2}$ or $r = 2$ . If $r = \dfrac{1}{2}$ then $b = \dfrac{34}{r} = 68$ and so $d = 25$ , i.e., an ascending arithmetic progression. If $r = 2$ then $b = 17$ and $d = -26$ , i.e., a descending arithmetic progression. Thus $r = \dfrac{1}{2}, d = 25, a = 42$ and $b = 68$ , giving us

$(a_{1}, a_{2}, a_{3}) = (17, 42, 67)$ and $(b_{1}, b_{2}, b_{3}) = (68, 34, 17)$ , and so

$a_{3} - a_{1} + b_{1} - b_{3} = 67 - 17 + 68 - 17 = \boxed{101}$ .