# The sum of three numbers in A.P. is 12 and the sum of their cubes is 408. Find the numbers.

Let the three numbers are in AP = a, a + d, a + 2d

According to the question,

The sum of three terms = 12

a + (a + d) + (a + 2d) = 12

3a + 3d = 12

a + d = 4

a = 4 – d …(i)

and the sum of their cubes = 408

a3 + (a + d)3 + (a + 2d)3 = 408

(4 – d)3 + (4)3 + ( 4 – d + 2d)3 = 408 [from(i)]

(4 – d)3 + (4)3 + ( 4 + d)3 = 408

64 – d3 + 12d2 – 48d + 64 + 64 + d3 + 12d2 + 48d = 408

192 + 24d2 = 408

24d2 = 408 – 192

24d2 = 216

d2 = 9

d = √9

d = ±3

Now, if d = 3, then a = 4 – 3 = 1

and if d = – 3, then a = 4 – ( – 3) = 4 + 3 = 7

So, the numbers are

if a = 1 and d = 3

1, 4, 7

and if a = 7 and d = – 3

7, 4, 1

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