The sum of three numbers in AP is 12. and the sum of their cubes is 288. Find the numbers.
Class 10
Arithmetic Progressions
Answers
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.