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If-the-sum-of-n-terms-f-a-series-is-A-n-2-B-n-where-A-B-are-constants-then-it-is-an-AP-

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Question Number 25741 by gugalesweta@gmail.com last updated on 13/Dec/17
If the sum of n terms f a series is A n^2 +B n , where A, B are constants, then it is an AP.
$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:{n}\:\mathrm{terms}\:\mathrm{f}\:\mathrm{a}\:\mathrm{series}\:\mathrm{is}\: \\ $$$${A}\:{n}^{\mathrm{2}} +{B}\:{n}\:,\:\mathrm{where}\:{A},\:{B}\:\mathrm{are}\:\mathrm{constants}, \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{is}\:\mathrm{an}\:\mathrm{AP}. \\ $$
Commented by math solver last updated on 13/Dec/17
yes.
$${yes}. \\ $$
Answered by Rasheed.Sindhi last updated on 14/Dec/17
S_n =An^2 +Bn S_(n−1) =A(n−1)^2 +B(n−1) T_n =(An^2 +Bn)−{A(n−1)^2 +B(n−1)} =An^2 −A(n−1)^2 +Bn−B(n−1) =An^2 −A(n^2 −2n+1)+Bn−Bn+B =2An−A+B The difference between consecutive terms d=T_k −T_(k−1) =2Ak−A+B−{2A(k−1)−A+B} =2Ak−A+B−2A(k−1)+A−B =2Ak−2Ak+2A =2A Since the difference between any two consecutive terms is constant The given sequence is AP.
$$\mathrm{S}_{\mathrm{n}} =\mathrm{An}^{\mathrm{2}} +\mathrm{Bn} \\ $$$$\mathrm{S}_{\mathrm{n}−\mathrm{1}} =\mathrm{A}\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{B}\left(\mathrm{n}−\mathrm{1}\right) \\ $$$$\mathrm{T}_{\mathrm{n}} =\left(\mathrm{An}^{\mathrm{2}} +\mathrm{Bn}\right)−\left\{\mathrm{A}\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{B}\left(\mathrm{n}−\mathrm{1}\right)\right\} \\ $$$$\:\:\:\:\:=\mathrm{An}^{\mathrm{2}} −\mathrm{A}\left(\mathrm{n}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{Bn}−\mathrm{B}\left(\mathrm{n}−\mathrm{1}\right) \\ $$$$\:\:\:=\mathrm{An}^{\mathrm{2}} −\mathrm{A}\left(\mathrm{n}^{\mathrm{2}} −\mathrm{2n}+\mathrm{1}\right)+\mathrm{Bn}−\mathrm{Bn}+\mathrm{B} \\ $$$$\:\:\:=\mathrm{2An}−\mathrm{A}+\mathrm{B} \\ $$$$\mathrm{The}\:\mathrm{difference}\:\mathrm{between}\:\mathrm{consecutive} \\ $$$$\mathrm{terms}\: \\ $$$$\mathrm{d}=\mathrm{T}_{\mathrm{k}} −\mathrm{T}_{\mathrm{k}−\mathrm{1}} =\mathrm{2Ak}−\mathrm{A}+\mathrm{B}−\left\{\mathrm{2A}\left(\mathrm{k}−\mathrm{1}\right)−\mathrm{A}+\mathrm{B}\right\} \\ $$$$\:\:\:=\mathrm{2Ak}−\mathrm{A}+\mathrm{B}−\mathrm{2A}\left(\mathrm{k}−\mathrm{1}\right)+\mathrm{A}−\mathrm{B} \\ $$$$\:\:\:=\mathrm{2Ak}−\mathrm{2Ak}+\mathrm{2A} \\ $$$$\:\:\:=\mathrm{2A} \\ $$$$\mathrm{Since}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{between}\:\mathrm{any} \\ $$$$\mathrm{two}\:\mathrm{consecutive}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{constant} \\ $$$$\mathrm{The}\:\mathrm{given}\:\mathrm{sequence}\:\mathrm{is}\:\mathrm{AP}. \\ $$
Commented by mrW1 last updated on 14/Dec/17
That′s great!
$${That}'{s}\:{great}! \\ $$

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