Question

Let $a,b,c$ and $d$ are in a geometric progression such that $a<b<c<d,a+d=112$ and $b+c=48.$ If the geometric progression is continued with $a$ as the first term, then the sum of the first six terms is

• A. $1156$
• B. $1256$
• C. $1356$
• D. $1456$

Answer 1

Answer: (D)

Let $r$ be the common ratio
$a+d=112\Rightarrow a+a{r}^3=112$
$b+c=48\Rightarrow ar+a{r}^2=48$
Dividing the first equation by the second equation we get
$\frac{a\left(1+r^3\right)}{a\left(1+r\right)r}=\frac{112}{48}=\frac{\left(1+r\right)\left(1-r+r^2\right)}{r\left(r+1\right)}=\frac{7}{3}$
$\Rightarrow 3r^2-3r+3=7r$
$\Rightarrow 3r^2-10r+3=0$
$r=3$ or $\frac{1}{3}$
Given they are in ascending order $\Rightarrow r=3$
$r=3\Rightarrow a=\frac{112}{1+r^3}=\frac{112}{28}=4$
$\Rightarrow a=4,b=12,c=36,d=108$
$\Rightarrow$ Sum of first $6$ terms $=\frac{4(3^6-1)}{(3-1)}=1456$