Question Number 202303 by hardmath last updated on 24/Dec/23
$$\mathrm{If}\:\:\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{1}\:\:\:\mathrm{and}\:\:\:\mathrm{a}_{\mathrm{1}} \:\centerdot\:\mathrm{a}_{\mathrm{2}} \:\centerdot\:…\:\centerdot\:\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:=\:\mathrm{n}^{\mathrm{2}} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a}_{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{13}} \:=\:? \\ $$
Answered by MATHEMATICSAM last updated on 24/Dec/23
![a_1 ∙ a_2 ∙ .... ∙ a_n = n^2 ∴ a_1 × a_2 × a_3 × .... × a_(13) = 13^2 = 169 ⇒ a_(13) = ((169)/(a_1 × a_2 × a_3 × ..... × a_(12) )) = ((169)/(144)) a_1 × a_2 = 2^2 = 4 ⇒ a_2 = 4 [∵ a_1 = 1] a_2 + a_(13) = 4 + ((169)/(144)) = ((745)/(144))](https://www.tinkutara.com/question/Q202309.png)
$${a}_{\mathrm{1}} \:\centerdot\:{a}_{\mathrm{2}} \:\centerdot\:….\:\centerdot\:{a}_{{n}} \:=\:{n}^{\mathrm{2}} \\ $$$$\therefore\:{a}_{\mathrm{1}} \:×\:{a}_{\mathrm{2}} \:×\:{a}_{\mathrm{3}} \:×\:….\:×\:{a}_{\mathrm{13}} \:=\:\:\mathrm{13}^{\mathrm{2}} \:=\:\mathrm{169} \\ $$$$\Rightarrow\:{a}_{\mathrm{13}} \:=\:\frac{\mathrm{169}}{{a}_{\mathrm{1}} \:×\:{a}_{\mathrm{2}} \:×\:{a}_{\mathrm{3}} \:×\:…..\:×\:{a}_{\mathrm{12}} }\:=\:\frac{\mathrm{169}}{\mathrm{144}} \\ $$$${a}_{\mathrm{1}} \:×\:{a}_{\mathrm{2}} \:=\:\mathrm{2}^{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\Rightarrow\:{a}_{\mathrm{2}} \:=\:\mathrm{4}\:\left[\because\:{a}_{\mathrm{1}} \:=\:\mathrm{1}\right] \\ $$$${a}_{\mathrm{2}} \:+\:{a}_{\mathrm{13}} \:=\:\mathrm{4}\:+\:\frac{\mathrm{169}}{\mathrm{144}}\:=\:\frac{\mathrm{745}}{\mathrm{144}} \\ $$
Commented by hardmath last updated on 24/Dec/23
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{ser} \\ $$
Answered by mr W last updated on 24/Dec/23
$${a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}−\mathrm{1}} {a}_{{n}} ={n}^{\mathrm{2}} \\ $$$${a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}−\mathrm{1}} =\left({n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} =\left(\frac{{n}}{{n}−\mathrm{1}}\right)^{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} +{a}_{\mathrm{13}} =\left(\frac{\mathrm{2}}{\mathrm{1}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{13}}{\mathrm{12}}\right)^{\mathrm{2}} =\frac{\mathrm{745}}{\mathrm{144}} \\ $$
Commented by hardmath last updated on 24/Dec/23
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{cool} \\ $$
Answered by 1990mbodji last updated on 24/Dec/23
$$ \\ $$$$\:\:{On}\:{suppose}\:{que}\:{a}_{\mathrm{1}} =\:\mathrm{1}\:\:{et}\:\:{a}_{\mathrm{1}} .{a}_{\mathrm{2}} …{a}_{{n}} \:=\:{n}^{\mathrm{2}} . \\ $$$${Trouvons}\:{la}\:{valeur}\:{de}\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{13}} . \\ $$$$\:\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} \:=\:\mathrm{4}\:\Rightarrow\:{a}_{\mathrm{2}} \:=\:\mathrm{4} \\ $$$$\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} {a}_{\mathrm{3}} \:=\:\mathrm{9}\:\Rightarrow\:{a}_{\mathrm{3}} \:=\:\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\:\:{Par}\:{it}\acute {{e}ration}\:\::\:{a}_{\mathrm{13}} \:=\:\frac{\mathrm{169}}{\mathrm{144}} \\ $$$$\:\:{Donc}\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{13}} \:=\:\mathrm{4}+\frac{\mathrm{169}}{\mathrm{144}}\:\:\Rightarrow\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{13}} \:=\:\frac{\mathrm{745}}{\mathrm{144}} \\ $$
Commented by hardmath last updated on 24/Dec/23
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{ser} \\ $$