Question

In a GP the sum of three numbers is $14,$ if $1$ is added to first two numbers and the third number is decreased by $1$, the series becomes AP, find the geometric sequence.

• A. $2,4,8$
• B. $8,4,2$
• C. $6,18,54$
• D. $8,16,32$

Answer 1

Answer: (A), (B)

$8,4,2$
$2,4,8$

$Letthenumbersana,ar,a{r}^{2}wherea=firsttermandr=commonratio\therefore a+ar+a{r}^2=14anda+1,ar+1,a{r}^2-1areinAP\therefore 2(ar+1)=(a+1)+(a{r}^2-1)or2ar+2=a+a{r}^{2}or3ar+2=a+ar+a{r}^{2}or3ar+2=14\therefore ar=4ora=\left(\frac{4}{r}\right)\therefore \left(\frac{4}{r}\right)+\left(\frac{4}{r}\right)\times r+\left(\frac{4}{r}\right)\times r^2=14or\left(\frac{4}{r}\right)+4+4r=14or4+4r+4r^2=14ror4r^2-10r+4=0or4r^2-8r-2r+4=0or4r(r-2)-2(r-2)=0or(4r-2)(r-2)=0\therefore r=\left(\frac{1}{2}\right)or2ifr=\left(\frac{1}{2}\right),thena=\left(\frac{4}{\left(\frac{1}{2}\right)}\right)=8\therefore sequence8,4,2andifr=2,thena=\left(\frac{4}{2}\right)=2\therefore sequence2,4,8$