find the consecutive term of AP whose sum is -3 and product of their cube is 512?
RaunakRaj:
it is the or three
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Let the three term be (a-d), a, (a+d)
So,
a-d+a+a+d= -3
3a= -3
a=(-1)
(a-d)^3×a^3×(a+d)^3=512
((a-d)(a+d))^3×a^3=512
((a^2-d^2)a)^3=512
((1-d^2)(-1))=8
1-d^2=(-8)
d^2=9
d=+3 or -3
So,
three terms are (-4), (-1) and 2.
PLZ MARK IT AS BRAINLIEST.
So,
a-d+a+a+d= -3
3a= -3
a=(-1)
(a-d)^3×a^3×(a+d)^3=512
((a-d)(a+d))^3×a^3=512
((a^2-d^2)a)^3=512
((1-d^2)(-1))=8
1-d^2=(-8)
d^2=9
d=+3 or -3
So,
three terms are (-4), (-1) and 2.
PLZ MARK IT AS BRAINLIEST.
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