Question

The sum of three numbers in AP is 12 and sum of their cubes is 288.

Chapter- 5 Arithmetic Progressions

Class 10

Answer 1

Answer:

Let the three terms be denoted as: ( a - d ), ( a ), ( a + d )

Sum of terms = 12 and Sum of cubes of terms = 288

=> a - d + a + a + d = 12

=> 3a = 12

=> a = 4

Also,

=> ( a - d )³ + ( a )³ + ( a + d )³ = 288

Substituting the value of 'a' we get,

=> ( 4 - d )³ + 4³ + ( 4 + d )³ = 288

=> ( 4³ - 3 ( 4 ) ( d ) ( 4 - d ) - d³ ) + 64 + ( 4 + d )³ = 288

=> ( 64 - 12 d ( 4 - d ) - d³ ) + 64 + ( 4 + d )³ = 288

=> ( 64 - 48d + 12d² - d³ ) + 64 + ( 4³ + 3 ( 4 ) ( d ) ( 4 + d ) + d³ ) = 288

=> ( 64 - 48d + 12d² - d³ ) + 64 + ( 64 + 12d ( 4 + d ) + d³ ) = 288

=> ( 64 - 48d + 12d² - d³ ) + 64 + ( 64 + 48d + 12d² + d³ ) = 288

=> 64 + 64 + 64 - 48d + 48d + 12d² + 12d² - d³ + d³ = 288

=> 192 + 24d² = 288

=> 24d² = 288 - 192

=> 24d² = 96

=> d² = 96 / 24 = 4

=> d = √4 = +2 or -2.

Now let us come to the AP.

Case 1: If a = 4 and d = +2, the AP would be:

=> 4 - 2, 4 , 4 + 2

=> 2, 4 , 6

Case 2: If a = 4 and d = -2, the AP would be:

=> 4 - ( -2 ) , 4 , 4 + ( -2 )

=> 4 + 2, 4, 4 - 2

=> 6, 4, 2

Hence the terms of the AP are 2, 4 and 6.