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An-exponential-sequence-of-positive-terms-and-a-linear-sequence-have-the-same-first-term-the-sum-o-their-first-term-is-3-the-sum-of-their-second-term-is-3-2-and-the-sum-of-their-third-term-is-6-

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Question Number 10488 by Saham last updated on 13/Feb/17
An exponential sequence of positive terms and a linear sequence have the same first term. the sum o their first term is 3, the sum of their second term is (3/2), and the sum of their third term is 6. find the sum of their fifth term.
$$\mathrm{An}\:\mathrm{exponential}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{terms}\:\mathrm{and}\:\mathrm{a} \\ $$$$\mathrm{linear}\:\mathrm{sequence}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{first}\:\mathrm{term}.\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{o}\:\mathrm{their}\:\mathrm{first}\:\mathrm{term}\:\mathrm{is}\:\mathrm{3},\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{second}\:\mathrm{term} \\ $$$$\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{third}\:\mathrm{term}\:\mathrm{is}\:\mathrm{6}.\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{fifth}\:\mathrm{term}. \\ $$
Answered by mrW1 last updated on 14/Feb/17
terms of A.P.: a_1 =a a_2 =a+d a_3 =a+2d a_4 =a+3d a_5 =a+4d ∙∙∙ terns of G.P.: b_1 =a b_2 =a×q b_3 =a×q^2 b_4 =a×q^3 b_5 =a×q^4 ∙∙∙ a_1 +b_1 =a+a=3 ⇒a=(3/2) a_2 +b_2 =a+d+a×q=(3/2) ⇒a+d+aq=(3/2) ⇒d+(3/2)q=0 ...(i) a_3 +b_3 =a+2d+a×q^2 =6 ⇒d+(3/4)q^2 =(9/4) ...(ii) (ii)−(i): (3/4)q^2 −(3/2)q−(9/4)=0 q^2 −2q−3=0 (q−3)(q+1)=0 ⇒q=3, since terms of G.P. are positive d=−(3/2)×3=−(9/2) a_5 =a+4d=(3/2)+4×(−(9/2))=−((33)/2) b_5 =a×q^4 =(3/2)×3^4 =((243)/2) a_5 +b_5 =−((33)/2)+((243)/2)=((210)/2)=105
$${terms}\:{of}\:{A}.{P}.: \\ $$$${a}_{\mathrm{1}} ={a} \\ $$$${a}_{\mathrm{2}} ={a}+{d} \\ $$$${a}_{\mathrm{3}} ={a}+\mathrm{2}{d} \\ $$$${a}_{\mathrm{4}} ={a}+\mathrm{3}{d} \\ $$$${a}_{\mathrm{5}} ={a}+\mathrm{4}{d} \\ $$$$\centerdot\centerdot\centerdot \\ $$$${terns}\:{of}\:{G}.{P}.: \\ $$$${b}_{\mathrm{1}} ={a} \\ $$$${b}_{\mathrm{2}} ={a}×{q} \\ $$$${b}_{\mathrm{3}} ={a}×{q}^{\mathrm{2}} \\ $$$${b}_{\mathrm{4}} ={a}×{q}^{\mathrm{3}} \\ $$$${b}_{\mathrm{5}} ={a}×{q}^{\mathrm{4}} \\ $$$$\centerdot\centerdot\centerdot \\ $$$${a}_{\mathrm{1}} +{b}_{\mathrm{1}} ={a}+{a}=\mathrm{3} \\ $$$$\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} +{b}_{\mathrm{2}} ={a}+{d}+{a}×{q}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{d}+{aq}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{d}+\frac{\mathrm{3}}{\mathrm{2}}{q}=\mathrm{0}\:\:\:\:\:…\left({i}\right) \\ $$$${a}_{\mathrm{3}} +{b}_{\mathrm{3}} ={a}+\mathrm{2}{d}+{a}×{q}^{\mathrm{2}} =\mathrm{6} \\ $$$$\Rightarrow{d}+\frac{\mathrm{3}}{\mathrm{4}}{q}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{4}}\:\:\:\:\:…\left({ii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{q}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{q}−\frac{\mathrm{9}}{\mathrm{4}}=\mathrm{0} \\ $$$${q}^{\mathrm{2}} −\mathrm{2}{q}−\mathrm{3}=\mathrm{0} \\ $$$$\left({q}−\mathrm{3}\right)\left({q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{q}=\mathrm{3},\:{since}\:{terms}\:{of}\:{G}.{P}.\:{are}\:{positive} \\ $$$${d}=−\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{3}=−\frac{\mathrm{9}}{\mathrm{2}} \\ $$$${a}_{\mathrm{5}} ={a}+\mathrm{4}{d}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{4}×\left(−\frac{\mathrm{9}}{\mathrm{2}}\right)=−\frac{\mathrm{33}}{\mathrm{2}} \\ $$$${b}_{\mathrm{5}} ={a}×{q}^{\mathrm{4}} =\frac{\mathrm{3}}{\mathrm{2}}×\mathrm{3}^{\mathrm{4}} =\frac{\mathrm{243}}{\mathrm{2}} \\ $$$${a}_{\mathrm{5}} +{b}_{\mathrm{5}} =−\frac{\mathrm{33}}{\mathrm{2}}+\frac{\mathrm{243}}{\mathrm{2}}=\frac{\mathrm{210}}{\mathrm{2}}=\mathrm{105} \\ $$
Commented by Saham last updated on 14/Feb/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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