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Question Number 187639 by LowLevelLump last updated on 19/Feb/23

Commented by a.lgnaoui last updated on 19/Feb/23

$$Thequestionislike$$$$v_{n+1}-v_n=a_{n+1}-a_n=\frac{1}{3}$$$$\frac{S_n}{a_n}=\mathrm{\Sigma }(v_n-u_n)?$$$$$$

Answered by Rasheed.Sindhi last updated on 20/Feb/23

$$A_n=\frac{S_n}{a_n}$$$$A_1=\frac{S_1}{a_1}=\frac{a}{a}=1;aisfirstterm$$$$A_2=\frac{a_1+a_2}{a_2}=\frac{a+a+d}{a+d}=\frac{a}{a+d}+1$$$$A_2-A_1=(\frac{a}{a+d}+1)-1=\frac{1}{1+d}=\frac{1}{3}$$$$\Rightarrow d=2$$$$\left(\mathrm{a}\right)Generalformulafor{a}_n$$$$a_n=a+(n-1)d=1+(n-1)\left(2\right)$$$$=1+2n-2=2n-1$$$$a_n=2n-1$$

Commented by mr W last updated on 20/Feb/23

$$a_n=2n-1$$$$S_n=\underset{k=1}{\sum ^n}(2k-1)=n^2$$$$A_n=\frac{S_n}{a_n}=\frac{n^2}{2n-1}$$$$clearly{A}_{n}isnotserieswithequal$$$$difference.$$$$A_1=1$$$$A_2=\frac{2^2}{2\times 2-1}=\frac{4}{3}\Rightarrow \frac{4}{3}-1=\frac{1}{3}$$$$A_3=\frac{3^2}{2\times 3-1}=\frac{9}{5}\Rightarrow \frac{9}{5}-\frac{4}{3}=\frac{7}{15}\ne \frac{1}{3}$$$$A_4=\frac{4^2}{2\times 4-1}=\frac{16}{7}\Rightarrow \frac{16}{7}-\frac{9}{5}=\frac{17}{35}\ne \frac{1}{3}$$$$......$$

Commented by mr W last updated on 20/Feb/23

$$ithinkthequestionissomewhere$$$$wrong.$$

Commented by Rasheed.Sindhi last updated on 20/Feb/23

$${You}^{\prime }reveryright,\mathcal{T}han\mathcal{X}\mathit{s}\mathit{i}\mathit{r}!$$

Answered by witcher3 last updated on 20/Feb/23

$$\frac{{\mathrm{s}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}+1}}-\frac{{\mathrm{s}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n}}}=\frac{1}{3};{\mathrm{a}}_1={\mathrm{s}}_1=1$$$${\mathrm{s}}_{\mathrm{n}+1}=\mathrm{a}$$$$\iff \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\frac{{\mathrm{s}}_{\mathrm{k}+1}}{{\mathrm{a}}_{\mathrm{k}+1}}-\frac{{\mathrm{s}}_{\mathrm{k}}}{{\mathrm{a}}_{\mathrm{k}}}=\frac{\mathrm{n}-1}{3},\mathrm{\forall }\mathrm{n}⩾2$$$$\frac{{\mathrm{s}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n}}}=\frac{\mathrm{n}+2}{3}$$$$\frac{{\mathrm{s}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}+1}}=\frac{\mathrm{n}+3}{3}\Rightarrow \frac{{\mathrm{s}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n}+1}}=\frac{\mathrm{n}}{3}$$$$\frac{{\mathrm{a}}_{\mathrm{n}+1}}{{\mathrm{a}}_{\mathrm{n}}}=\frac{\mathrm{n}+2}{\mathrm{n}}\Rightarrow \underset{\mathrm{k}=1}{\prod ^{\mathrm{n}-1}}\frac{{\mathrm{a}}_{\mathrm{k}+1}}{{\mathrm{a}}_{\mathrm{k}}}=\mathrm{\Pi }\frac{\mathrm{k}+2}{\mathrm{k}}=\frac{(\mathrm{n}+1)!}{(\mathrm{n}-1)!.2}=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$$${\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{2},{\mathrm{s}}_{\mathrm{n}}=\frac{\mathrm{n}+2}{3}{\mathrm{a}}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{6}$$$${\mathrm{s}}_{\mathrm{n}}=\frac{\mathrm{n}+2}{3}{\mathrm{a}}_{\mathrm{n}}$$$$\mathrm{\Sigma }\frac{1}{{\mathrm{a}}_{\mathrm{k}}}=\mathrm{\Sigma }\frac{2}{\mathrm{k}(\mathrm{k}+1)}=2\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}\frac{1}{\mathrm{k}}-\frac{1}{\mathrm{k}+1}=2(1-\frac{1}{\mathrm{n}+1})<2$$$$$$$$$$$$$$

Commented by mr W last updated on 20/Feb/23

$$thequestionsays\left\{a_n\right\}shouldbe$$$$seriesofequaldifference.but$$$$a_n=\frac{n(n+1)}{2}isnotserieswithequal$$$$difference.$$

Commented by witcher3 last updated on 20/Feb/23

$$\frac{{\mathrm{s}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n}}}\mathrm{is}\mathrm{arithmetic}\mathrm{equal}$$$$$$

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