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In a GP series consisting of positive terms, each term is equal to the sum of next two terms. Then, the common ratio of this GP series is
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Verified Answer
The correct answer is:
$\frac{\sqrt{5}-1}{2}$
Let $a_{n}$ be the general term of a GP whose first term is $a$ and common ratio is $r$. Now according to the question,
$$
\begin{aligned}
a_{n} &=a_{n+1}+a_{n+2} \\
a r^{n-1} &=a r^{n}+a r^{n+1}
\end{aligned}
$$
$\Rightarrow \quad r^{n-1}=r^{n}+r^{n+1}$
$\Rightarrow \quad 1=\frac{r^{n}}{r^{n-1}}+\frac{r^{n+1}}{r^{n-1}}$
$\Rightarrow \quad 1=r+r^{2}$
$\Rightarrow r^{2}+r-1=0$
$\Rightarrow \begin{aligned}r=& \frac{-1 \pm \sqrt{(1)^{2}-4(1)(-1)}}{2(1)} \\ &=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2} \end{aligned}$
since, GP consists only positive terms
$\therefore$
$r=\frac{\sqrt{5}-1}{2}$
$$
\begin{aligned}
a_{n} &=a_{n+1}+a_{n+2} \\
a r^{n-1} &=a r^{n}+a r^{n+1}
\end{aligned}
$$
$\Rightarrow \quad r^{n-1}=r^{n}+r^{n+1}$
$\Rightarrow \quad 1=\frac{r^{n}}{r^{n-1}}+\frac{r^{n+1}}{r^{n-1}}$
$\Rightarrow \quad 1=r+r^{2}$
$\Rightarrow r^{2}+r-1=0$
$\Rightarrow \begin{aligned}r=& \frac{-1 \pm \sqrt{(1)^{2}-4(1)(-1)}}{2(1)} \\ &=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2} \end{aligned}$
since, GP consists only positive terms
$\therefore$
$r=\frac{\sqrt{5}-1}{2}$
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