Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 46805 by 786786AM last updated on 31/Oct/18

In an A.p., the sum of first n terms is P , the sum of the next n terms is Q and   the sum of further next n terms is R. Show that P, Q, R is an A.P.

$$\mathrm{In}\:\mathrm{an}\:\mathrm{A}.\mathrm{p}.,\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{P}\:,\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{next}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{Q}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{further}\:\mathrm{next}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{R}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{P},\:\mathrm{Q},\:\mathrm{R}\:\mathrm{is}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18

first term is a  and common difference is d  P=(n/2)[2a+(n−1)d].....(1)  P+Q=((2n)/2)[2a+(2n−1)d]  P+Q+R=((3n)/2)[2a+(3n−1)d]  Q=((2n)/2)[2a+(2n−1)d]−(n/2)[2a+(n−1)d]       =(n/2)[4a+4nd−2d−2a−nd+d]        =(n/2)[2a+(3n−1)d]  Q−P  =(n/2)[2a+(3n−1)d−2a−(n−1)d]     =(n/2)[(3n−1−n+1)d     =n^2 d  R=P+Q+R−(P+Q)  R=((3n)/2)[2a+(3n−1)d]−((2n)/2)[{2a+(2n−1d)]  R=(n/2)[6a+9nd−3d−4a−4nd+2d]  R=(n/2)[2a+(5n−1)d]  R−Q  =(n/2)[2a+(5n−1)d]−(n/2)[2a+(3n−1)d]  =(n/2)[(5n−1−3n+1)d]  =(n/2)×2nd  =n^2 d  Q−P=R−Q  so P,Q,R  in A.P

$${first}\:{term}\:{is}\:{a}\:\:{and}\:{common}\:{difference}\:{is}\:{d} \\ $$$${P}=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right].....\left(\mathrm{1}\right) \\ $$$${P}+{Q}=\frac{\mathrm{2}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}\right){d}\right] \\ $$$${P}+{Q}+{R}=\frac{\mathrm{3}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right] \\ $$$${Q}=\frac{\mathrm{2}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}\right){d}\right]−\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left({n}−\mathrm{1}\right){d}\right] \\ $$$$\:\:\:\:\:=\frac{{n}}{\mathrm{2}}\left[\mathrm{4}{a}+\mathrm{4}{nd}−\mathrm{2}{d}−\mathrm{2}{a}−{nd}+{d}\right] \\ $$$$\:\:\:\:\:\:=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right] \\ $$$${Q}−{P} \\ $$$$=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}−\mathrm{2}{a}−\left({n}−\mathrm{1}\right){d}\right] \\ $$$$\:\:\:=\frac{{n}}{\mathrm{2}}\left[\left(\mathrm{3}{n}−\mathrm{1}−{n}+\mathrm{1}\right){d}\right. \\ $$$$\:\:\:={n}^{\mathrm{2}} {d} \\ $$$${R}={P}+{Q}+{R}−\left({P}+{Q}\right) \\ $$$${R}=\frac{\mathrm{3}{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right]−\frac{\mathrm{2}{n}}{\mathrm{2}}\left[\left\{\mathrm{2}{a}+\left(\mathrm{2}{n}−\mathrm{1}{d}\right)\right]\right. \\ $$$${R}=\frac{{n}}{\mathrm{2}}\left[\mathrm{6}{a}+\mathrm{9}{nd}−\mathrm{3}{d}−\mathrm{4}{a}−\mathrm{4}{nd}+\mathrm{2}{d}\right] \\ $$$${R}=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{5}{n}−\mathrm{1}\right){d}\right] \\ $$$${R}−{Q} \\ $$$$=\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{5}{n}−\mathrm{1}\right){d}\right]−\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{a}+\left(\mathrm{3}{n}−\mathrm{1}\right){d}\right] \\ $$$$=\frac{{n}}{\mathrm{2}}\left[\left(\mathrm{5}{n}−\mathrm{1}−\mathrm{3}{n}+\mathrm{1}\right){d}\right] \\ $$$$=\frac{{n}}{\mathrm{2}}×\mathrm{2}{nd} \\ $$$$={n}^{\mathrm{2}} {d} \\ $$$${Q}−{P}={R}−{Q} \\ $$$${so}\:{P},{Q},{R}\:\:{in}\:{A}.{P} \\ $$$$ \\ $$$$ \\ $$

Commented by 786786AM last updated on 01/Nov/18

Tk yu sir.

$$\mathrm{Tk}\:\mathrm{yu}\:\mathrm{sir}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18

most welcome...

$${most}\:{welcome}... \\ $$

Answered by MrW3 last updated on 01/Nov/18

let d=common difference  P=a_1 +a_2 +...+a_n   Q=a_(n+1) +a_(n+2) +...+a_(2n) =(a_1 +nd)+(a_2 +nd)+...+(a_n +nd)  =(a_1 +a_2 +...+a_n )+n×nd  =P+n^2 d  ⇒P−Q=n^2 d  similarly  R=a_(2n+1) +a_(2n+2) +...+a_(3n) =Q+n^2 d  ⇒R−Q=n^2 d    since R−Q=Q−P  ⇒P,Q,R are in A.P.

$${let}\:{d}={common}\:{difference} \\ $$$${P}={a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} \\ $$$${Q}={a}_{{n}+\mathrm{1}} +{a}_{{n}+\mathrm{2}} +...+{a}_{\mathrm{2}{n}} =\left({a}_{\mathrm{1}} +{nd}\right)+\left({a}_{\mathrm{2}} +{nd}\right)+...+\left({a}_{{n}} +{nd}\right) \\ $$$$=\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} \right)+{n}×{nd} \\ $$$$={P}+{n}^{\mathrm{2}} {d} \\ $$$$\Rightarrow{P}−{Q}={n}^{\mathrm{2}} {d} \\ $$$${similarly} \\ $$$${R}={a}_{\mathrm{2}{n}+\mathrm{1}} +{a}_{\mathrm{2}{n}+\mathrm{2}} +...+{a}_{\mathrm{3}{n}} ={Q}+{n}^{\mathrm{2}} {d} \\ $$$$\Rightarrow{R}−{Q}={n}^{\mathrm{2}} {d} \\ $$$$ \\ $$$${since}\:{R}−{Q}={Q}−{P} \\ $$$$\Rightarrow{P},{Q},{R}\:{are}\:{in}\:{A}.{P}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com