The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.
The sum of first three terms of a $G.P.$ is $\frac{13}{12}$ and their product is $-1$ Find the common ratio and the terms.
Let $\frac{a}{r}, a,$ ar be the first three terms of the $G.P.$ Then
$\frac{a}{r}+a r+a=\frac{13}{12}$ ........$(1)$
and $\left(\frac{a}{r}\right)(a)(a r)=-1$ ........$(2)$
From $(2),$ we get $a^{3}=-1,$ i.e., $a=-1$ (considering only real roots)
Substituting $a=-1$ in $(1),$ we have
$-\frac{1}{r}-1-r=\frac{13}{12}$ or $12 r^{2}+25 r+12=0$
This is a quadratic in $r$, solving, we get $r=-\frac{3}{4}$ or $-\frac{4}{3}$
Thus, the three terms of $G.P.$ are $: \frac{4}{3},-1, \frac{3}{4}$ for $r=\frac{-3}{4}$ and $\frac{3}{4},-1, \frac{4}{3}$ for $r=\frac{-4}{3}$
Similar Questions
Let $C_0$ be a circle of radius $I$ . For $n \geq 1$, let $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1} .$ Then, $\sum \limits_{i=0}^{\infty}$ Area $\left(C_i\right)$ equals
Let $C_0$ be a circle of radius $I$ . For $n \geq 1$, let $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1} .$ Then, $\sum \limits_{i=0}^{\infty}$ Area $\left(C_i\right)$ equals
- [KVPY 2014]
If $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$, then $x - a,\;y - a,\;z - a$ are in
If $2(y - a)$ is the $H.M.$ between $y - x$ and $y - z$, then $x - a,\;y - a,\;z - a$ are in
Find the sum up to $20$ terms in the geometric progression $0.15,0.015,0.0015........$
Find the sum up to $20$ terms in the geometric progression $0.15,0.015,0.0015........$
The value of $0.\mathop {234}\limits^{\,\,\, \bullet \,\, \bullet } $ is
The value of $0.\mathop {234}\limits^{\,\,\, \bullet \,\, \bullet } $ is
Three numbers are in $G.P.$ such that their sum is $38$ and their product is $1728$. The greatest number among them is
Three numbers are in $G.P.$ such that their sum is $38$ and their product is $1728$. The greatest number among them is