Menu Close

If-S-n-is-the-sum-of-the-first-n-terms-of-an-A-P-Express-S-2k-in-terms-of-S-k-and-S-3k-




Question Number 101779 by I want to learn more last updated on 04/Jul/20
If    S_n   is the sum of the first  n  terms of an A.P.  Express   S_(2k)   in terms of   S_k   and   S_(3k)
$$\mathrm{If}\:\:\:\:\mathrm{S}_{\mathrm{n}} \:\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\:\mathrm{n}\:\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$\mathrm{Express}\:\:\:\mathrm{S}_{\mathrm{2k}} \:\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\:\:\mathrm{S}_{\mathrm{k}} \:\:\mathrm{and}\:\:\:\mathrm{S}_{\mathrm{3k}} \\ $$
Answered by bemath last updated on 04/Jul/20
S_n  = ((n(a+u_n ))/2)→ S_(k ) = ((k(a+u_k ))/2)  u_k = ((2S_k )/k)−a ⇒a+(k−1)d=((2S_k )/k)−a  (k−1)d = ((2S_k )/k)−2a⇒d =((2S_k −2ak)/(k(k−1))) (•)  S_(2k)  = ((2k(a+u_(2k) ))/2) = k(a+u_(2k) )  u_(2k)  = (S_(2k) /k)−a ⇔a+(2k−1)d=(S_(2k) /k)−a   d = ((S_(2k) −ak)/(k(2k−1))) (••)  (•)=(••)  ((2S_k −2ak)/(k(k−1))) = ((S_(2k) −ak)/(k(2k−1)))  ⇒(((2k−1)(2S_k −2ak))/(k−1)) +ak = S_(2k)
$${S}_{\mathrm{n}} \:=\:\frac{\mathrm{n}\left(\mathrm{a}+\mathrm{u}_{\mathrm{n}} \right)}{\mathrm{2}}\rightarrow\:\mathrm{S}_{{k}\:} =\:\frac{{k}\left({a}+{u}_{{k}} \right)}{\mathrm{2}} \\ $$$${u}_{{k}} =\:\frac{\mathrm{2}{S}_{{k}} }{{k}}−{a}\:\Rightarrow{a}+\left({k}−\mathrm{1}\right){d}=\frac{\mathrm{2}{S}_{{k}} }{{k}}−{a} \\ $$$$\left({k}−\mathrm{1}\right){d}\:=\:\frac{\mathrm{2}{S}_{{k}} }{{k}}−\mathrm{2}{a}\Rightarrow{d}\:=\frac{\mathrm{2}{S}_{{k}} −\mathrm{2}{ak}}{{k}\left({k}−\mathrm{1}\right)}\:\left(\bullet\right) \\ $$$${S}_{\mathrm{2}{k}} \:=\:\frac{\mathrm{2}{k}\left({a}+{u}_{\mathrm{2}{k}} \right)}{\mathrm{2}}\:=\:{k}\left({a}+{u}_{\mathrm{2}{k}} \right) \\ $$$${u}_{\mathrm{2}{k}} \:=\:\frac{{S}_{\mathrm{2}{k}} }{{k}}−{a}\:\Leftrightarrow{a}+\left(\mathrm{2}{k}−\mathrm{1}\right){d}=\frac{{S}_{\mathrm{2}{k}} }{{k}}−{a}\: \\ $$$${d}\:=\:\frac{{S}_{\mathrm{2}{k}} −{ak}}{{k}\left(\mathrm{2}{k}−\mathrm{1}\right)}\:\left(\bullet\bullet\right) \\ $$$$\left(\bullet\right)=\left(\bullet\bullet\right) \\ $$$$\frac{\mathrm{2S}_{{k}} −\mathrm{2}{ak}}{{k}\left({k}−\mathrm{1}\right)}\:=\:\frac{{S}_{\mathrm{2}{k}} −{ak}}{{k}\left(\mathrm{2}{k}−\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{2}{S}_{{k}} −\mathrm{2}{ak}\right)}{{k}−\mathrm{1}}\:+{ak}\:=\:{S}_{\mathrm{2}{k}} \: \\ $$$$ \\ $$
Commented by I want to learn more last updated on 05/Jul/20
Thanks sir. I appreciate
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$

Leave a Reply

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