Question

The sum of the first three terms of a G.P. is $6$ and the sum of its first three odd terms is $10.5$. Then the sum of its first term and the common ratio can be

• A. $\frac{17}{12}$
• B. $\frac{15}{2}$
• C. $\frac{105}{38}$
• D. $\frac{115}{38}$

Answer 1

Answer: (B), (C)

$\frac{105}{38}$

Let the G.P. be $a,ar,a{r}^2,...$
Now, $a(1+r+r^2)=6...\left(1\right)$
$a(1+r^2+r^4)=10.5...\left(2\right)$
Dividing eqn $\left(2\right)$ by eqn $\left(1\right)$, we get
$\frac{a(1+r^2+r^4)}{a(1+r+r^2)}=\frac{10.5}{6}=\frac{7}{4}$
$\Rightarrow \frac{1+2r^2+r^4-r^2}{(1+r+r^2)}=\frac{7}{4}$
$\Rightarrow \frac{(1+r^2{)}^2-r^2}{(1+r+r^2)}=\frac{7}{4}$
$\Rightarrow \frac{(1+r+r^2)(1-r+r^2)}{(1+r+r^2)}=\frac{7}{4}$
$\Rightarrow 1-r+r^2=\frac{7}{4}$
$\Rightarrow 4r^2-4r-3=0$
$\Rightarrow (2r+1)(2r-3)=0$
$r=-\frac{1}{2}$ or $r=\frac{3}{2}$
If $r=-\frac{1}{2}\Rightarrow a=8$
If $r=\frac{3}{2}\Rightarrow a=\frac{24}{19}$
$\therefore a+r=\frac{15}{2}$ or $a+r=\frac{105}{38}$