Question

The sum of three numbers in G.P. is $42$. If first two of them are increased by $2$ and third decreased by $4$, then the polynomial whose zero are the numbers of G.P. is

• A. $x^3-42x^2-504x+1728$
• B. $x^3+42x^2+504x+1728$
• C. $x^3-42x^2+504x-1728$
• D. None of these

Answer 1

Answer: (C)

$x^3-42x^2+504x-1728$

Let the three number be $a,ar,a{r}^2,$ then

$a+ar+a{r}^2=42$ ...(1)

$2(ar+2)=(a+2)+(a{r}^2-4)\Rightarrow a-2ar+a{r}^2=6$ ...(2)

Subtracting (2) from (1), we have

$3ar=36\Rightarrow ar=12$ ...(3)

$\Rightarrow a+a{r}^2=30$ ...(4)

Solving (3) and (4). we get

$r=\frac{1}{2},2\Rightarrow a=24,6$

If $a=6$ and $r=\frac{1}{2}$ then $G.P.$ is $6,3,\frac{3}{2},\frac{3}{4}$

If $a=6$ and $r=2$ then $G.P.$ is $6,12,24,48$

If $a=24$ and $r=\frac{1}{2}$ then $G.P.$ is $24,12,6,3$

If $a=24$ and $r=2$ then $G.P.$ is $24,48,96,192$

Therefore from options only zero of $x^3-42x^2+504x-1728=0$ are from these $G.P{.}^{\prime }s$