Question

Four numbers are such that the first three are in A.P, while the last three in G.P. If the first number is 6 and common ratio of G.P. is $\frac{1}{2}$ then the numbers are.

• A. $6,9,12,6$
• B. $6,8,4,2$
• C. $6,10,14,7$
• D. $6,4,2,1$

Answer 1

Answer: (A), (D)

$6,9,12,6$
$6,4,2,1$

Let the four numbers be $a,a+d,a+2d,(a+d)r^2$.

where $d$ is the common difference of A.P. and $r$ is common ratio of the G.P.

$a=6$ are $r=\frac{1}{2}$ is given

$a+d,a+2d,(a+d)r^2$ are in G.P.

$\Rightarrow (a+2d{)}^2=(a+d{)}^2{r}^2\Rightarrow (6+2d{)}^2=(6+d{)}^2.\frac{1}{4}$

$\Rightarrow 4(6+2d{)}^2=(6+d{)}^2\Rightarrow 6+2d=(6+d).\frac{1}{2}$

$\Rightarrow d=-2\Rightarrow$ The $4$ numbers are $6,4,2,1$