Question

Let sum of first three terms of GP. with real terms is $\frac{13}{12}$ and their product is -1 . If the absolute value of the sum of their infinite terms is $\frac{p}{q}$ where $p$ and $q$ are coprime then $(q-p)$ is equal to

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (D)

Let $\frac{a}{r},a$, ar be three terms in GP.
$\therefore$ Product of terms $=a^3=-1$ (Given)
$\Rightarrow a=-1$
Now, sum of terms $=\frac{a}{r}+a+ar=\frac{13}{12}$ (Given)
$\Rightarrow \frac{-1}{r}-1-r=\frac{13}{12}$
$\Rightarrow 12r^2+25r+12=0$
$\therefore (3r+4)(4r+3)=0$
$\Rightarrow r=\frac{-4}{3},\frac{-3}{4}$
$\text{ But }r\ne \frac{-4}{3}\text{ (think?) }$
$\therefore \left|S_{\infty }\right|=\left|\frac{a/r}{1-r}\right|=\left|\frac{\frac{4}{3}}{1-\left(\frac{-3}{4}\right)}\right|=\left|\frac{\frac{4}{3}}{1+\frac{3}{4}}\right|=\left|\frac{16}{21}\right|=\frac{16}{21}=\frac{p}{q}$
$\Rightarrow p=16,q=21$
$\therefore q-p=21-16=5$