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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.

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

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