Question and Answers Forum

All Questions      Topic List

Arithmetic Questions

Previous in All Question      Next in All Question      

Previous in Arithmetic      Next in Arithmetic      

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.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com