Question-179482

Question Number 179482 by cherokeesay last updated on 29/Oct/22

Commented by mr W last updated on 30/Oct/22

$${it}\:{mustn}'{t}\:{be}\:{an}\:{arithmetic}\:{sequence}! \\ $$

Answered by mr W last updated on 29/Oct/22

a_n =s_n −s_(n−1) =2n+3n^2 −2(n−1)−3(n−1)^2 ⇒a_n =6n−1

$${a}_{{n}} ={s}_{{n}} −{s}_{{n}−\mathrm{1}} =\mathrm{2}{n}+\mathrm{3}{n}^{\mathrm{2}} −\mathrm{2}\left({n}−\mathrm{1}\right)−\mathrm{3}\left({n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}_{{n}} =\mathrm{6}{n}−\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 29/Oct/22

T_n =a+(n−1)d S_n =(n/2)[2a+(n−1)d]=2n+3n^2 2a+(n−1)d=(2/n)(2n+3n^2 )=4+6n 2a−d+nd=4+6n 2a−d=4 ∧ d=6 a=((4+d)/2)=((4+6)/2)=5 T_n =5+(n−1)(6)=6n−1

$${T}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d} \\ $$$${S}_{{n}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right]=\mathrm{2}{n}+\mathrm{3}{n}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}=\frac{\mathrm{2}}{{n}}\left(\mathrm{2}{n}+\mathrm{3}{n}^{\mathrm{2}} \right)=\mathrm{4}+\mathrm{6}{n} \\ $$$$\mathrm{2}{a}−{d}+{nd}=\mathrm{4}+\mathrm{6}{n} \\ $$$$\mathrm{2}{a}−{d}=\mathrm{4}\:\wedge\:{d}=\mathrm{6} \\ $$$$\:\:{a}=\frac{\mathrm{4}+{d}}{\mathrm{2}}=\frac{\mathrm{4}+\mathrm{6}}{\mathrm{2}}=\mathrm{5} \\ $$$${T}_{{n}} =\mathrm{5}+\left({n}−\mathrm{1}\right)\left(\mathrm{6}\right)=\mathrm{6}{n}−\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 30/Oct/22

determinant ((s_1 , ,s_2 , ,s_3 , ,s_4 , ,s_5 ,(...)),(( 5 _(a) ), ,(16), ,(33), ,(56), ,(85),(...)),( ,(11),( ),(17),( ),(23),( ),(29),( ),(...)),(( ), ,( 6 _(d) ), ,( 6),( ),( 6), ,( 6),(...))) T_n =a+(n−1)d=5+(n−1)(6) T_n =6n−1

$$\begin{array}{|c|c|c|c|}{\boldsymbol{{s}}_{\mathrm{1}} }&\hline{\:}&\hline{\boldsymbol{{s}}_{\mathrm{2}} }&\hline{\:}&\hline{\boldsymbol{{s}}_{\mathrm{3}} }&\hline{\:}&\hline{\boldsymbol{{s}}_{\mathrm{4}} }&\hline{\:}&\hline{\boldsymbol{{s}}_{\mathrm{5}} }&\hline{…}\\{\underset{\boldsymbol{{a}}} {\underbrace{\:\mathrm{5}\:}}}&\hline{\:}&\hline{\mathrm{16}}&\hline{\:}&\hline{\mathrm{33}}&\hline{\:}&\hline{\mathrm{56}}&\hline{\:}&\hline{\mathrm{85}}&\hline{…}\\{\:}&\hline{\mathrm{11}}&\hline{\:\:}&\hline{\mathrm{17}}&\hline{\:\:}&\hline{\mathrm{23}}&\hline{\:\:}&\hline{\mathrm{29}}&\hline{\:\:}&\hline{…}\\{\:\:}&\hline{\:}&\hline{\underset{\boldsymbol{{d}}} {\underbrace{\:\mathrm{6}\:}}}&\hline{\:}&\hline{\:\mathrm{6}}&\hline{\:\:}&\hline{\:\:\mathrm{6}}&\hline{\:}&\hline{\:\mathrm{6}}&\hline{…}\\\hline\end{array} \\ $$$$\boldsymbol{{T}}_{\boldsymbol{{n}}} =\boldsymbol{{a}}+\left(\boldsymbol{{n}}−\mathrm{1}\right)\boldsymbol{{d}}=\mathrm{5}+\left(\boldsymbol{{n}}−\mathrm{1}\right)\left(\mathrm{6}\right) \\ $$$$\boldsymbol{{T}}_{\boldsymbol{{n}}} =\mathrm{6}\boldsymbol{{n}}−\mathrm{1} \\ $$

Commented by Tawa11 last updated on 30/Oct/22

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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