The sum of three numbers in AP is 12 and sum of their cubes is 288.
Find the numbers.
Answers
Heyyy... here is ur answer
Let a-b,a and a+b be the three numbers
Given sum of the 3 numbers is 12
ie ,a-b+a+a+b = 12
3a =12
a =12÷3=4
Given sum of their cubes is 288
(a-b)^3+a^3+(a-b)^3 =288
a^3+b^3+3a^2b+3ab^2+a^3+a^3-b^3+3ab^2-3a^2b =288
Put a=4
After substituting a=4 we get the following eqn
64+12b^2+64+64+12b^2=288
192 +24b^2 =288
24b^2 =288-192=96
b^2 =96÷24=4
b =+/-2
Put b=2 or b=-2 in a-b,a and a +b
Hope this helps you
Marks as brainliest pls
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.