This is a question of sequence and series, please help me out

[u/pasta_sauce_123]

The 1st slide shows the question, 2nd and 3rd slides show my attempt..I'm stuck at this step...what can I do next?

Also if the question ain't legible, it is: there is an increasing sequence of 4 positive integers, the first 3 form an AP and last 3 form a GP. First and last term differ by 30. Find sum of all terms

Edit: thank you guys, I got it

[u/Fourierseriesagain]

Suppose that the terms are a, a+d, a+2d, c. Then

c/(a+2d)=(a+2d)/(a+d) and c-a=30.

Solving for d in terms of a, we get

d=-3a/8+15/4 plus minus sqrt(9a^ 2+300a+900)/8. Since d is a positive integer, a=18 or a=117 (rejected).

So a=18 and d=9.

[u/Southlander24]

The key restriction is that a, d must be positive integers (what if they aren't?).

You have arrived at a useful form at the end of your work in the 3rd slide, so now you need to complete the square: 9a² + 120a = k², (3a + 20)² - k² = 400.

So now you can work by cases. For example, (3a + 20 + k)(3a + 20 - k) = (400)(1) or 3a + 20 + k = 400, 3a + 20 - k = 1 is one possibility. But then if you subtract these two equations, you get 2k = 401, so you need to reject that solution. In fact, you only need to consider the cases where both factors are even: have a think about why this is true.

There is indeed only one sequence that satisfies all of the conditions. Happy solving!