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If $a_n=\sqrt{7+\sqrt{7+\sqrt{7+\ldots .}}}$ having $\mathrm{n}$ radical signs then by methods of mathematical induciton which is true
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$\mathrm{a}_{\mathrm{n}}>7 \forall \mathrm{n} \geq 1$
$\mathrm{a}_{\mathrm{n}}>7 \forall \mathrm{n} \geq 1$
$\mathrm{a}_1=\sqrt{7} < 7$. Leta $_{\mathrm{m}} < 7$. Then $\mathrm{a}_{\mathrm{m}}+1=\sqrt{7+\mathrm{a}_{\mathrm{m}}} \Rightarrow \mathrm{a}_{\mathrm{m}+1}^2=7+\mathrm{a}_{\mathrm{m}} < 7+7 < 14$
$\Rightarrow \mathrm{a}_{\mathrm{m}+1} < \sqrt{14} < 7 ;$ So $\mathrm{a}_{\mathrm{n}} < 7 \forall \mathrm{n} \therefore \mathrm{a}_{\mathrm{n}}>3$
$\Rightarrow \mathrm{a}_{\mathrm{m}+1} < \sqrt{14} < 7 ;$ So $\mathrm{a}_{\mathrm{n}} < 7 \forall \mathrm{n} \therefore \mathrm{a}_{\mathrm{n}}>3$
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