Let the three numbers in A.P. be a – d, a, a + d. Then (a – d) + a + (a + d) = 24 ⇒ 3a = 24 ⇒ a = 8 .....(i) and (a – d) a (a + d) = 440 ⇒ $(a^2 – d^2)$ a = 440 ⇒ $[8^2 – d^2]$ 8 = 440 [from (i)] ⇒ 64 – $d^2$ = 55 ⇒ $d^2$ = 64 – 55 ⇒ $d^2$ = 9 ⇒ d = ±3 Case (i) : If d = 3, then the numbers are 8 – 3, 8, 8 + 3 i.e., 5, 8, 11 Case (ii) : If d = – 3, then the numbers are 8 – ( – 3), 8, 8 + (– 3) i.e., 11, 8, 5