Find three consecutive terms in an A.P. whose sum is −3 and the product of their cubes is 512.
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×a3×(a+d)3=512
⇒(a2−d2)3×a3=512
⇒−(1−d2)3=512 (∵a=−1)
⇒−(1−d2)=8
⇒d2=9⇒d=±3
If d=3, then the numbers are −4,−1,2. If d=−3, then the numbers are 2,−1,−4.