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subtitute $a + c = 2b$ into the 1st equation $a + b + c = 60 \Rightarrow a + c + b = 60 \Rightarrow 2b + b = 60 \Rightarrow 3b = 60 \Rightarrow b = 20$ so we get b = 20. Subtitute b = 20 into the 1st equation $a + 20 + c = 60 \Rightarrow a + c = 40 \Rightarrow a = 40-c$

subtitute $b = 20$ and $a = 40 - c$ into the 3rd equation $20^2 = (40 - c) (c + 10)$ $400 = 40c + 400 - c^2 - 10c$ $c^2 - 30c = 0$ $c (c-30) = 0$

$c1 = 0 \Rightarrow a1 = 40$ $c2 = 30 \Rightarrow a2 = 10$ The answer which satifies $c > a$ is $c = 30$ and $a = 10$

$c - a = 30 - 10 = 20$