Official solution
Let the A.P is
a - 2d, a - d, a, a + d, a + 2d
\[ \because \] sum = 25
\[ \Rightarrow \] 5a = 25 \[ \Rightarrow \] a = 5
Also given,
product (a2 -- 4d2) (a2 -- d2).a = 2520
\[ \Rightarrow \] (25 -- 4d2) (25 --d2)5 = 2520
\[ \Rightarrow \] 4d4
-- 121d2
-- 4d2
+ 121 = 0
\[ \Rightarrow \] (d2
-- 1) (4d2 -- 121) = 0
\[ \Rightarrow \] d = \[ \pm \]1, d = \[ \pm {{11} \over 2}\]
When d = \[ \pm \]1 we can't get any fraction term like -\[{1 \over 2}\].
\[ \therefore \] d = \[ \pm {{11} \over 2}\]
And when d = \[{{11} \over 2}\]
we get largest term = 5 + 2d = 5 + 11 = 16