Find three number in AP such that sum is 12 and product is 408
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Here's , ur answer :-
Solution =)
=) 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.
______________________
HOPE , IT HELPS ... ✌️
_______________________
_______________________
Here's , ur answer :-
Solution =)
=) 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.
______________________
HOPE , IT HELPS ... ✌️
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