Question

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

Answer 1

Answer:

The Numbers are 2, 4 and 6 or  6, 4 and 2.

Step-by-step explanation:

Given :  

Sum of three terms of an A.P. is 12

Let the three terms in AP are (a - d), a, (a + d)

Then, we have

(a – d) + a + (a + d) = 12

3a = 12

a = 12/3

a = 4 ………..(1)

Also it is given that the, sum of their cubes is 288 , therefore we have,  

(a – d)³ + a³ + (a + d)³ = 288

a³ – d³ – 3ad (a – d) + a³ + a³ + d³ + 3ad (a + d) = 288

[By using the formula , (a – b)³ = a³ – b³ – 3ab (a – b) ]

a³ – d³ – 3a²d + 3ad² + a³ + a³ + d³ + 3a²d + 3ad² = 288

3a³ – 3a²d + 3ad² + 3a²d + 3ad² = 288

3a³ + 6ad² + = 288

3(4)³ + 6(4)d² = 288

[From eq 1]

3 × 64 + 24d² = 288

192 + 24d² = 288

24d² = 288 - 192

24d² = 96

d² = 96/24

d² = 4

d = √4

d = ±4

Case 1 :  

when a = 4 and d = 2, then ,  

First number, (a – d) = 4 – 2 = 2

Third number ,(a + d) = 4 + 2 = 6

Therefore, the numbers are 2, 4 and 6  

Case 2 :  

When a = 4 and d = -2, then ,

First number, (a – d) = 4 + 2 = 6

Third number ,(a + d) = 4 – 2 = 2

Therefore, the numbers are 6, 4 and 2

Hence, the Numbers are 2, 4 and 6 or  6, 4 and 2.

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Answer 2

Step-by-step explanation:

Solution.................