Question

If the sum of three numbers in A.P. is 12 and sum of their cubes is 408, find the numbers.
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Answer 1

Answer :-

The numbers are :- 1, 4 and 7

Explanation :-

Since the numbers are in A.P. let the three numbers be (a - d), a, (a + d).

Given, their sum is 12

$\sf \implies a -d + a + a +d = 12\\ \\\implies 3a = 12\\\\\implies a = \frac{12}{3}\\\\\implies a = 4$

So, we got that second number, which is a is 4

Now, sum of their cubes is 408

$\sf \implies (a - d)^3 + a^3 + (a + d)^3 = 408\\\\\implies \tiny{a^3 - d^3 - 3a^2d + 3ad^2 + a^3 + a^3 + d^3 + 3a^2d + 3ad^2 = 408}\\\\\implies 3a^3 + 6ad^2 = 408 \\\\Now \ substitute \ the \ value \ of \ a \ as \ 4\\\\\implies 3(4)^3 + 6(4)d^2 = 408\\\\\implies 3(64) + 24d^2 = 408\\\\\implies 192 + 24d^2 = 408\\\\\implies 24d^2 = 408 - 192\\\\\implies d^2 = 216/24\\\\\implies d^2 = 9\\\\\implies d = \sqrt{9}\\\\\implies d = \pm 3$

Here, we will take d as 3 only, because negative value would not satisfy the condition.

So, taking d as 3, we get the first number :-

(a - d) = (4 - 3) = 1

second number :- a = 4

third number :- (a + 3) = (4 + 3) = 7

$\rule{200}2$