Question
Find the next three tearms of the sequence :
$\sqrt{5}, 5,5 \sqrt{5}............$
$\sqrt{5}, 5,5 \sqrt{5}............$
✓
Answer
Given sequence : $\sqrt{5}, 5,5 \sqrt{5}$..
Now,
$\frac{5}{\sqrt{5}}=\sqrt{5}, \frac{5 \sqrt{5}}{5}=\sqrt{5}$
Since $\frac{5}{\sqrt{5}}=\frac{5 \sqrt{5}}{52}=\ldots \ldots \ldots . .=\sqrt{5}$,
the given sequence is a G.P. with first term, $a=\sqrt{5}$ and
Now, $t_n = ar^{n-1}$
$\therefore$ Next three term:
$4^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^3=\sqrt{5} \times 5 \sqrt{5}=25$
$5^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^4=\sqrt{5} \times 25=25 \sqrt{5}$
$6^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^5=\sqrt{5} \times 25 \sqrt{5}=125$
Now,
$\frac{5}{\sqrt{5}}=\sqrt{5}, \frac{5 \sqrt{5}}{5}=\sqrt{5}$
Since $\frac{5}{\sqrt{5}}=\frac{5 \sqrt{5}}{52}=\ldots \ldots \ldots . .=\sqrt{5}$,
the given sequence is a G.P. with first term, $a=\sqrt{5}$ and
Now, $t_n = ar^{n-1}$
$\therefore$ Next three term:
$4^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^3=\sqrt{5} \times 5 \sqrt{5}=25$
$5^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^4=\sqrt{5} \times 25=25 \sqrt{5}$
$6^{\text {th }} \text { term }=\sqrt{5} \times(\sqrt{5})^5=\sqrt{5} \times 25 \sqrt{5}=125$
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