Question Number 17187 by virus last updated on 02/Jul/17
Answered by Tinkutara last updated on 02/Jul/17
![((a_1 + a_2 + ... + a_p )/(a_1 + a_2 + ... + a_q )) = (((p/2)[2a_1 + (p − 1)d])/((q/2)[2a_1 + (q − 1)d])) = ((p[2a_1 + (p − 1)d])/(q[2a_1 + (q − 1)d])) = (p^2 /q^2 ) (Given) 2a_1 q + (p − 1)dq = 2a_1 p + (q − 1)dp 2a_1 (p − q) = d(pq − q − pq + p) 2a_1 = d ⇒ a_1 = (d/2) Now (a_6 /a_(12) ) = ((a_1 + 5d)/(a_1 + 11d)) = (((11d)/2)/((23d)/2)) = ((11)/(23))](https://www.tinkutara.com/question/Q17188.png)
$$\frac{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{p}} }{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{2}} \:+\:…\:+\:{a}_{{q}} }\:=\:\frac{\frac{{p}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({p}\:−\:\mathrm{1}\right){d}\right]}{\frac{{q}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({q}\:−\:\mathrm{1}\right){d}\right]} \\ $$$$=\:\frac{{p}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({p}\:−\:\mathrm{1}\right){d}\right]}{{q}\left[\mathrm{2}{a}_{\mathrm{1}} \:+\:\left({q}\:−\:\mathrm{1}\right){d}\right]}\:=\:\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:\left(\mathrm{Given}\right) \\ $$$$\mathrm{2}{a}_{\mathrm{1}} {q}\:+\:\left({p}\:−\:\mathrm{1}\right){dq}\:=\:\mathrm{2}{a}_{\mathrm{1}} {p}\:+\:\left({q}\:−\:\mathrm{1}\right){dp} \\ $$$$\mathrm{2}{a}_{\mathrm{1}} \left({p}\:−\:{q}\right)\:=\:{d}\left({pq}\:−\:{q}\:−\:{pq}\:+\:{p}\right) \\ $$$$\mathrm{2}{a}_{\mathrm{1}} \:=\:{d}\:\Rightarrow\:{a}_{\mathrm{1}} \:=\:\frac{{d}}{\mathrm{2}} \\ $$$$\mathrm{Now}\:\frac{{a}_{\mathrm{6}} }{{a}_{\mathrm{12}} }\:=\:\frac{{a}_{\mathrm{1}} \:+\:\mathrm{5}{d}}{{a}_{\mathrm{1}} \:+\:\mathrm{11}{d}}\:=\:\frac{\frac{\mathrm{11}{d}}{\mathrm{2}}}{\frac{\mathrm{23}{d}}{\mathrm{2}}}\:=\:\frac{\mathrm{11}}{\mathrm{23}} \\ $$
Answered by RasheedSoomro last updated on 03/Jul/17
![AnOther way ((a_1 +a_2 +...a_p )/(a_1 +a_2 +...a_q ))=(p^2 /q^2 ) , (a_6 /a_(12) )=? [We require ((a_1 +5d)/(a_1 +11d))] (((p/2)[2a_1 +(p−1)d])/((q/2)[2a_1 +(q−1)d]))=(p^2 /q^2 ) ((a_1 +(((p−1)/2))d)/(a_1 +(((q−1)/2))d))=(p/q) Now let ((p−1)/2)=5⇒p=11 and ((q−1)/2)=11⇒q=23 So (a_6 /a_(12) )=((a_1 +5d)/(a_1 +11d))=((11)/(23))](https://www.tinkutara.com/question/Q17251.png)
$$\mathrm{AnOther}\:\mathrm{way} \\ $$$$\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…{a}_{{p}} }{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +…{a}_{{q}} }=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:\:\:,\:\:\:\:\frac{{a}_{\mathrm{6}} }{{a}_{\mathrm{12}} }=? \\ $$$$\left[{W}\mathrm{e}\:\mathrm{require}\:\frac{{a}_{\mathrm{1}} +\mathrm{5}{d}}{{a}_{\mathrm{1}} +\mathrm{11}{d}}\right] \\ $$$$\:\:\:\frac{\frac{{p}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} +\left({p}−\mathrm{1}\right){d}\right]}{\frac{{q}}{\mathrm{2}}\left[\mathrm{2}{a}_{\mathrm{1}} +\left({q}−\mathrm{1}\right){d}\right]}=\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} } \\ $$$$\frac{{a}_{\mathrm{1}} +\left(\frac{{p}−\mathrm{1}}{\mathrm{2}}\right){d}}{{a}_{\mathrm{1}} +\left(\frac{{q}−\mathrm{1}}{\mathrm{2}}\right){d}}=\frac{{p}}{{q}} \\ $$$$\mathrm{Now}\:\mathrm{let}\:\frac{{p}−\mathrm{1}}{\mathrm{2}}=\mathrm{5}\Rightarrow{p}=\mathrm{11} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{and}\:\frac{{q}−\mathrm{1}}{\mathrm{2}}=\mathrm{11}\Rightarrow{q}=\mathrm{23} \\ $$$${S}\mathrm{o} \\ $$$$\frac{{a}_{\mathrm{6}} }{{a}_{\mathrm{12}} }=\frac{{a}_{\mathrm{1}} +\mathrm{5}{d}}{{a}_{\mathrm{1}} +\mathrm{11}{d}}=\frac{\mathrm{11}}{\mathrm{23}} \\ $$
Answered by myintkhaing last updated on 03/Jul/17
$$\mathrm{another}\:\mathrm{way} \\ $$$$\mathrm{Let}\:\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +\mathrm{a}_{\mathrm{3}} +…+\mathrm{a}_{\mathrm{p}} =\mathrm{S}_{\mathrm{p}} =\mathrm{kp}^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +\mathrm{a}_{\mathrm{3}} +…+\mathrm{a}_{\mathrm{q}} =\mathrm{S}_{\mathrm{q}} =\mathrm{kq}^{\mathrm{2}} \\ $$$$\mathrm{So},\:\mathrm{S}_{\mathrm{n}} =\:\mathrm{kn}^{\mathrm{2}} \\ $$$$\mathrm{Thus} \\ $$$$\mathrm{a}_{\mathrm{6}} =\:\mathrm{S}_{\mathrm{6}} −\mathrm{S}_{\mathrm{5}} =\mathrm{11k}\:\mathrm{and}\:\mathrm{a}_{\mathrm{12}} =\mathrm{S}_{\mathrm{12}} −\mathrm{S}_{\mathrm{11}} =\mathrm{23k} \\ $$$$\therefore\:\frac{\mathrm{a}_{\mathrm{6}} }{\mathrm{a}_{\mathrm{12}} }\:=\:\frac{\mathrm{11}}{\mathrm{23}} \\ $$
Commented by virus last updated on 15/Jul/17
$${yes}\:{thank}\:{you} \\ $$
Commented by RasheedSoomro last updated on 03/Jul/17
$$\mathrm{More}\:\mathrm{efficient}\:\mathrm{way}\:\mathrm{than}\:\mathrm{that}\:\mathrm{of}\:\mathrm{mine}! \\ $$