Question

Find three consecutive terms in A.P. whose sum is – 3 and the
product of their cubes is 512. (Take the value of d positive.)​

Answer 1

Answer:

Let the numbers be (a−d),a,(a+d). Then,

Sum=−3⇒(a−d)+a+(a+d)=−3⇒3a=−3⇒a=−1

Product of their cubes=512⇒(a−d) 3 ×a 3 ×(a+d)3 =512⇒(a 2 −d 2 ) 3×a 3 =512

⇒−(1−d 2 ) 3=512 (∵a=−1)

⇒−(1−d 2)=8 ⇒d 2

=9⇒d=±3

If d=3, then the numbers are −4,−1,2. If d=−3, then the numbers are 2,−1,−4.

Answer 2

Answer:

-4,-1 and 2

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