Question 53
An arithmetic progression is such that the 12th term is 42. What should the common difference be, such that the product of the first, second and 33rd terms is maximum?
Solution
Let the first term of the series be a and the common difference be d.
The 12th term is a+11d = 42. Sum of the first, second and 33rd terms = a+(a+d)+(a+32d) = 3a + 33d = 3(a+11d) = 3*42 = 126.
Product of the first, second and 33rd terms is a*(a+d)*(a+32d).
Since the sum is constant, the product is maximum when all the terms are equal. => a = a+d = a + 32d => d = 0.
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