Three terms are in arithmetic progression such that their sum is 36 and product is 1296. Find the three terms.a.4, 12, 20b.9, 12, 15c.8, 12, 16d.6, 12, 18
Question
Solution
The three terms in an arithmetic progression can be represented as (a-d), a, (a+d), where a is the middle term and d is the common difference.
Given that the sum of the three terms is 36, we can write the equation:
(a-d) + a + (a+d) = 36 3a = 36 a = 36/3 a = 12
The product of the three terms is given as 1296, so we can write the equation:
(a-d)a(a+d) = 1296 12*(12-d)(12+d) = 1296 (12-d)(12+d) = 1296/12 (12-d)*(12+d) = 108 144 - d^2 = 108 d^2 = 144 - 108 d^2 = 36 d = sqrt(36) d = 6
So, the three terms are (12-6), 12, (12+6) or 6, 12, 18. Therefore, the correct answer is d.6, 12, 18.
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