A particle starts its motion from rest and moves with constant acceleration for time $t_1$ and then it retards with constant rate for $t_2$. And comes to rest. Then the ratio of maximum speed and average speed during the complete motion will be
MY SOLUTION
Let acceleration be a.
Max speed
$$v=at_1$$
Also distance covered will be 
$$s=\frac{(2)(at_1^2)}{2}$$
$$=at_1^2$$
So average speed 
$$=\frac{at_1^2}{t_1+t_2}$$
Taking their ratio gives 
$$\frac{t_1+t_2}{t_1}$$
That’s as far as I got. But the answer is 2:1 and I have no idea on how to get there. Please help me proceed. 
Thanks!

You have assumed that the magnitude of the acceleration and deceleration is equal, which is not given. 
For the acceleration phase, $v_{max}=a_1t_1$ and distance covered is $\frac12a_1t_1^2$.
For the deceleration phase, $v^2-u^2=-a_1^2t_1^2=-2a_2s$ giving the distance covered as $\frac{a_1^2t_1^2}{2a_2}$. You also have $v=0=a_1t_1-a_2t_2$. Thus the average speed is $$\frac12\frac{a_1t_1^2+\frac{a_1^2t_1^2}{a_2}}{t_1+t_2}=\frac12\frac{a_1t_1^2+a_2t_2^2}{t_1+t_2}$$Taking the ratio$$\frac{v_{avg}}{v_{max}}=\frac12\frac{\frac{a_1t_1^2}{a_1t_1}+\frac{a_2t_2^2}{a_1t_1}}{t_1+t_2}=1/2$$
