a.p

• First, lets assume " A 50" as $\alpha$ , " A 100" as $\beta$ and " A 1" as $\theta$ , and we have to remember : $\theta + 49r = \alpha$ .
• We have:
  $50\alpha = 100\beta => \alpha = 2\beta$ .
  
• We also have: $\beta = \alpha + 50r$ , which we can use to obtain: $\alpha = 2\alpha + 100r => \alpha = -100r$ .
• Now, using $\alpha = \theta + 49r$ , we have: $-100r = \theta + 49r => 0 = \theta + 149r =$ A 150.
• Concluding, we have A 150 = 0.
  PS: i am sorry i used many terms and letters, solution guide didn't help me at all. If you have any question to ask, i'm all yours to help you understand this problem.
  

We write the general term of an a.p as ( a+n-1*d) where n denotes the term we are finding a relationship for. Writing so for the above equation we get 100(a+99d)= 50(a+49d). Solving we get, a=-149 d. Thus 150th tern= 0 :)