The third,sixth and seventh terms of a geometric progression(whose common ratio is neither 0 nor 1) are in arithmetic progression. Prove dat d sum of d first three is equal to d fourth term

Answer 1

ar^5 - ar^2 = 3(ar^6-ar^5)
r^3-1 = 3r^4 - 3r^3
3r^4 - 4r^3 + 1 = 0
The only real root of this equation is r=1.

Is there a typo? Can you find an error I made?