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
$a+d=112\Rightarrow a+ar^{3}=112$
$b+c=48\Rightarrow ar+ar^{2}=48$
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 $=\frac{4 \left(\right. 3^{6} - 1 \left.\right)}{\left(\right. 3 - 1 \left.\right)}=1456$