If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.
Let (a - d), a, (a + d) be three numbers in A.P.
∴(a−d)+a+(a+d)=24
⇒3a=24⇒a=8
and (a−d)(a)(a+d)=440……(1)
Substituting a = 8 in (1), we have:
(8−d)(8)(8+d)=440
⇒(64−d2)8=440
⇒64−d2=4408=55
⇒d2=64−55=9=32
⇒d±3
When d = + 3, then the A.P is (8 - 3), 8, (8 + 3) or 5, 8, 11
when d = - 3 then the A.P. is
[8+(−3)],8,[8+(−3)]
or 8 + 3, 8, 8 - 3
or 11, 8, 5