find three number whose sum is 12 and sum of their cubes is 408
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Given sum of three numbers is 12
Let the terms be, a, a+d, a+2d. Given that sum is 12.
That is, a+(a+d)+(a+2d) = 12
That is, 3a+3d = 12,
Meaning, a+d=4
Since sum is 12, numbers can be (1,4,7) or (2,4,6) or (3,4,5).
But it's given that sum of cubes is 408. This is satisfied only by (1,4,7).
Hence the numbers are 1,4,7.
Let the terms be, a, a+d, a+2d. Given that sum is 12.
That is, a+(a+d)+(a+2d) = 12
That is, 3a+3d = 12,
Meaning, a+d=4
Since sum is 12, numbers can be (1,4,7) or (2,4,6) or (3,4,5).
But it's given that sum of cubes is 408. This is satisfied only by (1,4,7).
Hence the numbers are 1,4,7.
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