Question

If the sum of the first three terms of a G.P. is $\frac{13}{12}$ and their product is $-1,$ then which of the following can be terms of those G.P.?

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $-1$
• D. $1$

Answer 1

Answer: (A), (B), (C)

$-1$

Let the first three terms of the given G.P. be $\frac{a}{r},a$ and $ar$.
Then, $\frac{a}{r}\times a\times ar=-1$
$\Rightarrow a^3=-1\Rightarrow a=-1$
And, $\frac{a}{r}+a+ar=\frac{13}{12}$
$\Rightarrow \frac{-1}{r}-1-r=\frac{13}{12}(\because a=-1)$
$\Rightarrow \frac{-1}{r}-r=\frac{13}{12}+1$
$\Rightarrow \frac{-1-r^2}{r}=\frac{25}{12}$
$\Rightarrow 12r^2+25r+12=0$
$\Rightarrow (3r+4)(4r+3)=0$
$\Rightarrow r=-\frac{4}{3}$ or $r=-\frac{3}{4}$

So, the required terms of G.P. are
$\frac{3}{4},-1,\frac{4}{3}$ or $\frac{4}{3},-1,\frac{3}{4}$