The Catchy AP

Let the first terms and common differences of the two arithmetic progressions be $a_1$ and $d_1$ , and $a_2$ and $d_2$ respectively. Then,

\(\begin{array} {} n = 1: & \dfrac {a_1}{a_2} = \dfrac {1-1}{1+1} = \dfrac 01 & \implies a_1 = 0 \\ n = 2: & \dfrac {d_1}{2a_2+d_2} = \dfrac 13 & \implies 3d_1 = 2a_2+d_2 & ...(1) \\ n = 3: & \dfrac {2d_1}{2a_2+2d_2} = \dfrac 12 & \implies 4d_1 = 2a_2+2d_2 & ...(2) \\ (2) - (1): & d_1 = d_2 \\ (1): & 3d_1 = 2a_2+d_1 & \implies d_1 = a_2 \end{array} \)

Therefore, the ratio of the fifth terms of the two APs is $\dfrac {a_1+4d_1}{a_2+4d_2} = \dfrac {4a_2}{a_2+4a_2} = \dfrac 45$ .

$\implies a+b = 4+5=\boxed{9}$

Let us denote the first term of the first AP as $x$ and the ratio as $R$ , and for the second AP, the first term as $y$ and ratio as $S$ . Sum on n first terms for the first AP is:

$2x + (n-1)R$

And for the second AP is:

$2y + (n-1)S$

The problem says that:

$\frac{2x + (n-1)R}{2y + (n-1)S} = \frac{n-1}{n+1}$

Or:

$\frac{x + \frac{(n-1)R}{2}}{y + \frac{(n-1)S}{2}} = \frac{n-1}{n+1}$ .

The ratio of the fifth terms is:

$\frac{x +4R}{y + 4S}$

Which is the same of plugging $\frac{n-1}{2}$ as $4$ , or $n$ as $9$ in the equation above. This leads to $\frac{8}{10}$ or $\frac{4}{5}$ . So, $a = 4, b = 5$ and $a + b = \fbox{9}$