Question Number 173926 by mnjuly1970 last updated on 21/Jul/22
$$ \\ $$$$\:\:\:\:\:{Q}: \\ $$$$\:\:\:\:\:\:\:{a}_{\:{n}\:} \:\:{is}\:{an}\:{arithmatic}\:{sequence}. \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}\:\left(\:{first}\:{term}\right)\:{and}\:\:{d}\:\left({difference}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{such}\:{that}\:,\:\:{a}_{\:{a}} \:+\:{a}_{\:{d}} \:=\:{a}_{\:{ad}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{find}\:\:\::\:\:\:\:{a}_{\:{n}} \:\:=? \\ $$$$\:\:\:\:\:\:\:\:\:\:{note}:\:\:{a}\:\:,\:\:{d}\:\:\:\in\:\mathbb{N}\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$
Commented by Rasheed.Sindhi last updated on 21/Jul/22
$$\:{a}\:\left(\:{first}\:{term}\right)\:{and}\:\:{d}\:\left({differece}\right) \\ $$
Commented by mnjuly1970 last updated on 21/Jul/22
$$\:\:{grateful}\:{sir}\:{Rashid} \\ $$
Commented by Rasheed.Sindhi last updated on 21/Jul/22
$$\:\:\mathbb{T}\boldsymbol{\mathrm{han}}\mathbb{X}\:{mnjuly}\:{sir}! \\ $$
Answered by mahdipoor last updated on 21/Jul/22
$$\Rightarrow\left[{a}+\left({a}−\mathrm{1}\right){d}\right]+\left[{a}+\left({d}−\mathrm{1}\right){d}\right]=\left[{a}+\left({ad}−\mathrm{1}\right){d}\right] \\ $$$$\Rightarrow{a}+{ad}+{d}^{\mathrm{2}} −{d}={ad}^{\mathrm{2}} \\ $$$$\Rightarrow{a}\left(\mathrm{1}+{d}−{d}^{\mathrm{2}} \right)={d}−{d}^{\mathrm{2}} \Rightarrow{a}=\frac{{d}−{d}^{\mathrm{2}} }{{d}−{d}^{\mathrm{2}} −\mathrm{1}}\in\mathrm{N}\Rightarrow \\ $$$$\begin{cases}{{d}−{d}^{\mathrm{2}} ={ma}}\\{{d}−{d}^{\mathrm{2}} −\mathrm{1}={m}\in\mathrm{Z}}\end{cases}\Rightarrow\mathrm{1}={m}\left({a}−\mathrm{1}\right)\Rightarrow \\ $$$$\begin{cases}{{m}=−\mathrm{1},{a}=\mathrm{0}\:\Rightarrow{d}−{d}^{\mathrm{2}} \:=\mathrm{0}\Rightarrow{d}=\mathrm{0},\mathrm{1}}\\{{m}=\mathrm{1},{a}=\mathrm{2}\:\Rightarrow{d}−{d}^{\mathrm{2}} =\mathrm{2}\Rightarrow\nexists{d}\in\mathrm{N}}\end{cases} \\ $$$$\Rightarrow\Rightarrow\:\begin{cases}{{ans}\:\mathrm{1}:\:{a}_{{n}} =\mathrm{0}\:\:\:\:\:\:\:\:\:{with}\:\:{a}=\mathrm{0}\:,\:{d}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:}\\{{ans}\:\mathrm{2}:\:{a}_{{n}} ={n}−\mathrm{1}\:\:{with}\:\:{a}=\mathrm{0}\:,\:{d}=\mathrm{1}}\end{cases} \\ $$
Commented by Tawa11 last updated on 21/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by Rasheed.Sindhi last updated on 21/Jul/22
$${a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d} \\ $$$${a}_{{a}} ={a}+\left({a}−\mathrm{1}\right){d}\:\:,\:\:{a}_{{d}} ={a}+\left({d}−\mathrm{1}\right){d}, \\ $$$${a}_{{ad}} ={a}+\left({ad}−\mathrm{1}\right){d} \\ $$$$ \\ $$$${a}_{{a}} +{a}_{{d}} ={a}_{{ad}} \\ $$$$\Rightarrow{a}+\left({a}−\mathrm{1}\right){d}+{a}+\left({d}−\mathrm{1}\right){d} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={a}+\left({ad}−\mathrm{1}\right){d} \\ $$$$\Rightarrow\mathrm{2}{a}+\left({a}+{d}−\mathrm{2}\right){d}={a}+\left({ad}−\mathrm{1}\right){d} \\ $$$$\:\:\:\:\:\:{a}+\left({a}+{d}−\mathrm{2}−{ad}+\mathrm{1}\right){d}=\mathrm{0} \\ $$$$\:\:\:\:\:\:{a}+\left({a}+{d}−{ad}−\mathrm{1}\right){d}=\mathrm{0} \\ $$$$\:\:\:\:\:{a}+{ad}+{d}^{\mathrm{2}} −{ad}^{\mathrm{2}} −{d}=\mathrm{0} \\ $$$${a}\left(\mathrm{1}+{d}−{d}^{\mathrm{2}} \right)=−{d}^{\mathrm{2}} +{d} \\ $$$${a}=\frac{{d}^{\mathrm{2}} −{d}}{{d}^{\mathrm{2}} −{d}−\mathrm{1}}\in\mathbb{N}\Rightarrow{d}^{\mathrm{2}} −{d}−\mathrm{1}\:\mid\:{d}^{\mathrm{2}} −{d} \\ $$$${d}^{\mathrm{2}} −{d}=\mathrm{0}\:\vee\:{d}^{\mathrm{2}} −{d}−\mathrm{1}=\mathrm{1} \\ $$$${d}\left({d}−\mathrm{1}\right)=\mathrm{0}\:\vee\:{d}\left({d}−\mathrm{1}\right)=\mathrm{2} \\ $$$${d}=\overset{×} {\mathrm{0}},\mathrm{1},\mathrm{2}\Rightarrow{a}=\overset{×} {\mathrm{0}},\overset{×} {\mathrm{0}},\mathrm{2} \\ $$$$\Rightarrow{d}=\mathrm{2}\Rightarrow{a}=\mathrm{2}\:\checkmark \\ $$$${a}_{{n}} =\mathrm{2}+\left({n}−\mathrm{1}\right)\left(\mathrm{2}\right)=\mathrm{2}+\mathrm{2}{n}−\mathrm{2}=\mathrm{2}{n} \\ $$$$\begin{array}{|c|}{{a}_{{n}} =\mathrm{2}{n}}\\\hline\end{array} \\ $$$$\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8},…\:\:\: \\ $$$${Verification}: \\ $$$${a}_{{a}} ={a}+\left({a}−\mathrm{1}\right){d}\:{or}\:{a}_{{d}} ={a}+\left({d}−\mathrm{1}\right){d} \\ $$$${a}_{\mathrm{2}} =\mathrm{2}+\left(\mathrm{2}−\mathrm{1}\right)\left(\mathrm{2}\right)=\mathrm{4} \\ $$$${a}_{{ad}} ={a}+\left({ad}−\mathrm{1}\right){d} \\ $$$${a}_{\mathrm{2}×\mathrm{2}} =\mathrm{2}+\mathrm{3}×\mathrm{2}=\mathrm{8} \\ $$$${a}_{\mathrm{4}} =\mathrm{8} \\ $$$${a}_{{a}} +{a}_{{d}} ={a}_{{ad}} \\ $$$${a}_{\mathrm{2}} +{a}_{\mathrm{2}} ={a}_{\mathrm{4}} \\ $$$$\mathrm{4}+\mathrm{4}=\mathrm{8} \\ $$
Commented by mnjuly1970 last updated on 21/Jul/22
$${thanks}\:{alot}\:{sir} \\ $$