Math, asked by NainaMehra, 1 year ago

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

Answered by Steph0303
15

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.



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