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Divide 144 into 3 parts which are in arithmetic progression (AP) such that the largest is twice the smallest.

What is the smallest of the three numbers?

A. 48
B. 36
C. 13
D. 32

Answer :

Final answer:

The smallest number of the three, when divided proportionately into an arithmetic progression where the total is 144 and the largest is twice the smallest, is (b) 36.

Explanation:

Given the problem, we know that the numbers are in an arithmetic progression or AP, with three terms, and that they total 144. An arithmetic progression is a sequence of numbers in which the difference of any two successive members is a constant.

According to the problem, the largest number is twice the smallest, meaning the progression is something like 'a', 'a + d', '2a', where 'a' is the smallest number and 'd' is the common difference.

From the progression given, combined with the fact that the total of the numbers is 144, we can form the equation a + (a + d) + 2a = 144. Simplifying this equation gets us 4a + d = 144.

The set of possible smallest numbers provided only include option b. 36, which satisfies this equation when d = 36. Hence, the smallest number is 36.

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