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} $
$\Rightarrow 12 r^2+25 r+12=0$
$\therefore (3 r+4)(4 r+3)=0 $
$\Rightarrow r=\frac{-4}{3}, \frac{-3}{4} $
$\text { But } r \neq \frac{-4}{3} \text { (think?) }$
$\therefore \left| S _{\infty}\right|=\left|\frac{ a / r }{1- r }\right|=\left|\frac{\frac{4}{3}}{1-\left(\frac{-3}{4}\right)}\right|=\left|\frac{\frac{4}{3}}{1+\frac{3}{4}}\right|=\left|\frac{16}{21}\right|=\frac{16}{21}=\frac{ p }{ q }$
$\Rightarrow p=16, q=21 $
$\therefore q-p=21-16=5$