Let a1, a2, a3, a4 be an A.P. of four terms such that each term of the A.P. and its common difference l are integers. If a1 + a2 + a3 + a4 = 48 and a1a2a3a4 + l4 = 361, then the largest term of the A.P. is equal to
- 27
- 24
- 23
- 21
Answer
(1) 27
Solution
a1 + a4 = a2 + a3 = 24 and a3 - a2 = l, so a2 = 12 - l2.
Since a2 is an integer, l is even. Putting l = 2k (k integer), the terms are
12 - 3k, 12 - k, 12 + k, 12 + 3k
and the given condition becomes
(144 - 9k2)(144 - k2) + 16k4 = 361
20736 - 1440k2 + 25k4 = 361
5k4 - 288k2 + 4075 = 0
k2 = 288 3810 = 25 or 32.6
Since k is an integer, k2 = 25, i.e. k = 5.
∴ The terms are -3, 7, 17, 27 (or in reverse order) and the largest term = 27.
Answer: (1)