the sum of three number in a.p.is 12 and the sum of their cubes is 288. find the numbers.
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Answers
Answered by
5
Let the numbers be: a, (a-d), (a+d).
Sum of the numbers is: a+a-d+a+d=3a
Given sum of numbers is 12.
Therefore, a=4.
Next, a^3+(a+d)^3+(a-d)^3=288.
As given in the question.
So,
a^3+d^3+3ad(a+d)+a^3-d^3–3ad(a-d)=224.
= a^3+3ad(a+d)-3ad(a-d)=224.
= 2a(a^2+3d^2)=224.
= a^2+3d^2=28.
= 16+3d^2=28.
= d^2=4.
= d=2.
Therefore, a=4 and d=2.
Thus, the 3 numbers are: 2, 4, 6.
Sum of the numbers is: a+a-d+a+d=3a
Given sum of numbers is 12.
Therefore, a=4.
Next, a^3+(a+d)^3+(a-d)^3=288.
As given in the question.
So,
a^3+d^3+3ad(a+d)+a^3-d^3–3ad(a-d)=224.
= a^3+3ad(a+d)-3ad(a-d)=224.
= 2a(a^2+3d^2)=224.
= a^2+3d^2=28.
= 16+3d^2=28.
= d^2=4.
= d=2.
Therefore, a=4 and d=2.
Thus, the 3 numbers are: 2, 4, 6.
Hi Megha
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Answered by
1
Answer:
2,4,6
Step-by-step explanation:
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