If a_1 = 1 and a_1 ∙ a_2 ∙ ... ∙ a_n = n^2 Find: a_2 + a


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Question Number 202303 by hardmath last updated on 24/Dec/23

$If a_{1} = 1 and a_{1} \cdot a_{2} \cdot . . . \cdot a_{n} = n^{2}$ $Find : a_{2} + a_{13} = ?$
	Answered by MATHEMATICSAM last updated on 24/Dec/23

	$a_{1} \cdot a_{2} \cdot . . . . \cdot a_{n} = n^{2}$ $∴ a_{1} \times a_{2} \times a_{3} \times . . . . \times a_{13} = 13^{2} = 169$ $\Rightarrow a_{13} = \frac{169}{a_{1} \times a_{2} \times a_{3} \times . . . . . \times a_{12}} = \frac{169}{144}$ $a_{1} \times a_{2} = 2^{2} = 4$ $\Rightarrow a_{2} = 4 [∵ a_{1} = 1]$ $a_{2} + a_{13} = 4 + \frac{169}{144} = \frac{745}{144}$
	Commented by hardmath last updated on 24/Dec/23

	$thankyou dear ser$
	Answered by mr W last updated on 24/Dec/23

	$a_{1} a_{2} . . . a_{n - 1} a_{n} = n^{2}$ $a_{1} a_{2} . . . a_{n - 1} = {(n - 1)}^{2}$ $\Rightarrow a_{n} = {(\frac{n}{n - 1})}^{2}$ $a_{2} + a_{13} = {(\frac{2}{1})}^{2} + {(\frac{13}{12})}^{2} = \frac{745}{144}$
	Commented by hardmath last updated on 24/Dec/23

	$thankyou dear professor cool$
	Answered by 1990mbodji last updated on 24/Dec/23

	$O n s u p p o s e q u e a_{1} = 1 e t a_{1} . a_{2} . . . a_{n} = n^{2} .$ $T r o u v o n s l a v a l e u r d e a_{2} + a_{13} .$ $a_{1} a_{2} = 4 \Rightarrow a_{2} = 4$ $a_{1} a_{2} a_{3} = 9 \Rightarrow a_{3} = \frac{9}{4}$ $P a r i t \overset{´}{e r a t i o n} : a_{13} = \frac{169}{144}$ $D o n c a_{2} + a_{13} = 4 + \frac{169}{144} \Rightarrow a_{2} + a_{13} = \frac{745}{144}$
	Commented by hardmath last updated on 24/Dec/23

	$thankyou dear ser$
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