Menu Close

Question-187639

Tinku Tara 0 Comments
Pin




Question Number 187639 by LowLevelLump last updated on 19/Feb/23
Commented by a.lgnaoui last updated on 19/Feb/23
The question is like v_(n+1) −v_n =a_(n+1) −a_n =(1/3) (S_n /a_n )=Σ(v_n −u_n )?
$${The}\:{question}\:{is}\:{like} \\ $$$${v}_{{n}+\mathrm{1}} −{v}_{{n}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\frac{{S}_{{n}} }{{a}_{{n}} }=\Sigma\left({v}_{{n}} −{u}_{{n}} \right)? \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 20/Feb/23
A_n =(S_n /a_n ) A_1 =(S_1 /a_1 )=(a/a)=1; a is first term A_2 =((a_1 +a_2 )/a_2 )=((a+a+d)/(a+d))=(a/(a+d))+1 A_2 −A_1 =((a/(a+d))+1)−1=(1/(1+d))=(1/3) ⇒d=2 (a)General formula for a_n a_n =a+(n−1)d=1+(n−1)(2) =1+2n−2=2n−1 a_n =2n−1
$${A}_{{n}} =\frac{{S}_{{n}} }{{a}_{{n}} } \\ $$$${A}_{\mathrm{1}} =\frac{{S}_{\mathrm{1}} }{{a}_{\mathrm{1}} }=\frac{{a}}{{a}}=\mathrm{1};\:{a}\:{is}\:{first}\:{term} \\ $$$${A}_{\mathrm{2}} =\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} }{{a}_{\mathrm{2}} }=\frac{{a}+{a}+{d}}{{a}+{d}}=\frac{{a}}{{a}+{d}}+\mathrm{1} \\ $$$${A}_{\mathrm{2}} −{A}_{\mathrm{1}} =\left(\frac{{a}}{{a}+{d}}+\mathrm{1}\right)−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}+{d}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow{d}=\mathrm{2} \\ $$$$\left(\mathrm{a}\right){General}\:{formula}\:{for}\:{a}_{{n}} \\ $$$$\:\:{a}_{{n}} ={a}+\left({n}−\mathrm{1}\right){d}=\mathrm{1}+\left({n}−\mathrm{1}\right)\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{2}{n}−\mathrm{2}=\mathrm{2}{n}−\mathrm{1} \\ $$$$\:\:{a}_{{n}} =\mathrm{2}{n}−\mathrm{1} \\ $$
Commented by mr W last updated on 20/Feb/23
a_n =2n−1 S_n =Σ_(k=1) ^n (2k−1)=n^2 A_n =(S_n /a_n )=(n^2 /(2n−1)) clearly A_n is not series with equal difference. A_1 =1 A_2 =(2^2 /(2×2−1))=(4/3) ⇒(4/3)−1=(1/3) A_3 =(3^2 /(2×3−1))=(9/5) ⇒(9/5)−(4/3)=(7/(15))≠(1/3) A_4 =(4^2 /(2×4−1))=((16)/7) ⇒((16)/7)−(9/5)=((17)/(35))≠(1/3) ......
$${a}_{{n}} =\mathrm{2}{n}−\mathrm{1} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{2}{k}−\mathrm{1}\right)={n}^{\mathrm{2}} \\ $$$${A}_{{n}} =\frac{{S}_{{n}} }{{a}_{{n}} }=\frac{{n}^{\mathrm{2}} }{\mathrm{2}{n}−\mathrm{1}} \\ $$$${clearly}\:{A}_{{n}} \:{is}\:{not}\:{series}\:{with}\:{equal} \\ $$$${difference}. \\ $$$${A}_{\mathrm{1}} =\mathrm{1} \\ $$$${A}_{\mathrm{2}} =\frac{\mathrm{2}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}−\mathrm{1}}=\frac{\mathrm{4}}{\mathrm{3}}\:\:\Rightarrow\frac{\mathrm{4}}{\mathrm{3}}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${A}_{\mathrm{3}} =\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}−\mathrm{1}}=\frac{\mathrm{9}}{\mathrm{5}}\:\Rightarrow\frac{\mathrm{9}}{\mathrm{5}}−\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{15}}\neq\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${A}_{\mathrm{4}} =\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}×\mathrm{4}−\mathrm{1}}=\frac{\mathrm{16}}{\mathrm{7}}\:\Rightarrow\frac{\mathrm{16}}{\mathrm{7}}−\frac{\mathrm{9}}{\mathrm{5}}=\frac{\mathrm{17}}{\mathrm{35}}\neq\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$…… \\ $$
Commented by mr W last updated on 20/Feb/23
i think the question is somewhere wrong.
$${i}\:{think}\:{the}\:{question}\:{is}\:{somewhere} \\ $$$${wrong}. \\ $$
Commented by Rasheed.Sindhi last updated on 20/Feb/23
You′re very right, ThanX sir!
$${You}'{re}\:{very}\:{right},\:\mathcal{T}{han}\mathcal{X}\:\boldsymbol{{sir}}! \\ $$
Answered by witcher3 last updated on 20/Feb/23
(s_(n+1) /a_(n+1) )−(s_n /a_n )=(1/3);a_1 =s_1 =1 s_(n+1) =a ⇔Σ_(k=1) ^(n−1) (s_(k+1) /a_(k+1) )−(s_k /a_k )=((n−1)/3),∀n≥2 (s_n /a_n )=((n+2)/3) (s_(n+1) /a_(n+1) )=((n+3)/3)⇒(s_n /a_(n+1) )=(n/3) (a_(n+1) /a_n )=((n+2)/n)⇒Π_(k=1) ^(n−1) (a_(k+1) /a_k )=Π((k+2)/k)=(((n+1)!)/((n−1)!.2))=((n(n+1))/2) a_n =((n(n+1))/2),s_n =((n+2)/3)a_n =((n(n+1)(n+2))/6) s_n =((n+2)/3)a_n Σ(1/a_k )=Σ(2/(k(k+1)))=2Σ_(k=1) ^n (1/k)−(1/(k+1))=2(1−(1/(n+1)))<2
$$\frac{\mathrm{s}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }−\frac{\mathrm{s}_{\mathrm{n}} }{\mathrm{a}_{\mathrm{n}} }=\frac{\mathrm{1}}{\mathrm{3}};\mathrm{a}_{\mathrm{1}} =\mathrm{s}_{\mathrm{1}} =\mathrm{1} \\ $$$$\mathrm{s}_{\mathrm{n}+\mathrm{1}} =\mathrm{a} \\ $$$$\Leftrightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{s}_{\mathrm{k}+\mathrm{1}} }{\mathrm{a}_{\mathrm{k}+\mathrm{1}} }−\frac{\mathrm{s}_{\mathrm{k}} }{\mathrm{a}_{\mathrm{k}} }=\frac{\mathrm{n}−\mathrm{1}}{\mathrm{3}},\forall\mathrm{n}\geqslant\mathrm{2} \\ $$$$\frac{\mathrm{s}_{\mathrm{n}} }{\mathrm{a}_{\mathrm{n}} }=\frac{\mathrm{n}+\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{\mathrm{s}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }=\frac{\mathrm{n}+\mathrm{3}}{\mathrm{3}}\Rightarrow\frac{\mathrm{s}_{\mathrm{n}} }{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }=\frac{\mathrm{n}}{\mathrm{3}} \\ $$$$\frac{\mathrm{a}_{\mathrm{n}+\mathrm{1}} }{\mathrm{a}_{\mathrm{n}} }=\frac{\mathrm{n}+\mathrm{2}}{\mathrm{n}}\Rightarrow\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\frac{\mathrm{a}_{\mathrm{k}+\mathrm{1}} }{\mathrm{a}_{\mathrm{k}} }=\Pi\frac{\mathrm{k}+\mathrm{2}}{\mathrm{k}}=\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\left(\mathrm{n}−\mathrm{1}\right)!.\mathrm{2}}=\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}},\mathrm{s}_{\mathrm{n}} =\frac{\mathrm{n}+\mathrm{2}}{\mathrm{3}}\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}{\mathrm{6}} \\ $$$$\mathrm{s}_{\mathrm{n}} =\frac{\mathrm{n}+\mathrm{2}}{\mathrm{3}}\mathrm{a}_{\mathrm{n}} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{k}} }=\Sigma\frac{\mathrm{2}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}=\mathrm{2}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{1}}=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\right)<\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 20/Feb/23
the question says {a_n } should be series of equal difference. but a_n =((n(n+1))/2) is not series with equal difference.
$${the}\:{question}\:{says}\:\left\{{a}_{{n}} \right\}\:{should}\:{be} \\ $$$${series}\:{of}\:{equal}\:{difference}.\:{but} \\ $$$${a}_{{n}} =\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}\:{is}\:{not}\:{series}\:{with}\:{equal} \\ $$$${difference}. \\ $$
Commented by witcher3 last updated on 20/Feb/23
(s_n /a_n ) is arithmetic equal
$$\frac{\mathrm{s}_{\mathrm{n}} }{\mathrm{a}_{\mathrm{n}} }\:\mathrm{is}\:\mathrm{arithmetic}\:\mathrm{equal} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *