The sum of three numbers in A.P. is 12 and the sum of their cubes is 4

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

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

Answer»

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