The sum of first three terms of a $G.P.$ is $\frac{39}{10}$ and their product is $1 .$ Find the common ratio and the terms.
The sum of first three terms of a $G.P.$ is $\frac{39}{10}$ 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 $G.P.$
$\frac{a}{r}+a+a r=\frac{39}{10}$ ..........$(1)$
$\left(\frac{a}{r}\right)(a)(a r)=1$ .........$(2)$
From $(2),$ we Obtain $a^{3}=1$
$\Rightarrow a=1$ (Considering real roots only)
Substituting $a=1$ in equation $(1),$ we obtain
$\frac{1}{r}+1+r=\frac{39}{10}$
$\Rightarrow 1+r+r^{2}=\frac{39}{10} r$
$\Rightarrow 10+10 r+10 r^{2}-39 r=0$
$\Rightarrow 10 r^{2}-29 r+10=0$
$\Rightarrow 10 r^{2}-25 r-4 r+10=0$
$\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$
$\Rightarrow(5 r-2)(2 r-5)=0$
$\Rightarrow r=\frac{2}{5}$ or $\frac{5}{2}$
Thus, the three terms of $G.P.$ are $\frac{5}{2}, 1$ and $\frac{2}{5}$
Similar Questions
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.
The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is
The number of natural number $n$ in the interval $[1005, 2010]$ for which the polynomial. $1+x+x^2+x^3+\ldots+x^{n-1}$ divides the polynomial $1+x^2+x^4+x^6+\ldots+x^{2010}$ is
- [KVPY 2010]
Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a $G.P.$ Prove that $P ^{2} R ^{n}= S ^{n}$
Let $S$ be the sum, $P$ the product and $R$ the sum of reciprocals of $n$ terms in a $G.P.$ Prove that $P ^{2} R ^{n}= S ^{n}$
For $0<\mathrm{c}<\mathrm{b}<\mathrm{a}$, let $(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^2+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}$ $+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements
$(I)$ If $\alpha \in(-1,0)$, then $\mathrm{b}$ cannot be the geometric mean of $\mathrm{a}$ and $\mathrm{c}$
$(II)$ If $\alpha \in(0,1)$, then $\mathrm{b}$ may be the geometric mean of $a$ and $c$
For $0<\mathrm{c}<\mathrm{b}<\mathrm{a}$, let $(\mathrm{a}+\mathrm{b}-2 \mathrm{c}) \mathrm{x}^2+(\mathrm{b}+\mathrm{c}-2 \mathrm{a}) \mathrm{x}$ $+(c+a-2 b)=0$ and $\alpha \neq 1$ be one of its root. Then, among the two statements
$(I)$ If $\alpha \in(-1,0)$, then $\mathrm{b}$ cannot be the geometric mean of $\mathrm{a}$ and $\mathrm{c}$
$(II)$ If $\alpha \in(0,1)$, then $\mathrm{b}$ may be the geometric mean of $a$ and $c$
- [JEE MAIN 2024]
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 $0.15,0.015,0.0015........$