Question

Find three consecutive terms in an A.P. whose sum is $-3$ and the product of their cubes is $512$.

• A. $-4,-1,2$
• B. $4,1,2$
• C. $-4,2,3$
• D. None of these

Answer 1

The correct option is A $-4,-1,2$

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

$Sum=-3\Rightarrow (a-d)+a+(a+d)=-3\Rightarrow 3a=-3\Rightarrow a=-1$

Product of their cubes$=512\Rightarrow (a-d{)}^3\times a^3\times (a+d{)}^3=512$

$\Rightarrow (a^2-d^2{)}^3\times a^3=512$

$\Rightarrow -(1-d^2{)}^3=512$ ($\because a=-1$)

$\Rightarrow -(1-d^2)=8$

$\Rightarrow d^2=9\Rightarrow d=\pm 3$

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