Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 175829 by pete last updated on 07/Sep/22

The 2^(nd) term of a Geometric Progresion (G.P) is equal to the 8^(th) term of an Arithmetic Progresion (A.P). The first terms, common difference and common ratio are all equal and non−zero. Find the sum of the first five terms of the Geometric Progresion(G.P)

$$\mathrm{The}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Geometric}\:\mathrm{Progresion} \\ $$$$\left(\mathrm{G}.\mathrm{P}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{Arithmetic} \\ $$$$\mathrm{Progresion}\:\left(\mathrm{A}.\mathrm{P}\right).\:\mathrm{The}\:\mathrm{first}\:\mathrm{terms},\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{and}\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{are}\:\mathrm{all}\:\mathrm{equal} \\ $$$$\mathrm{and}\:\mathrm{non}−\mathrm{zero}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{five} \\ $$$$\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Geometric}\:\mathrm{Progresion}\left(\mathrm{G}.\mathrm{P}\right) \\ $$

Answered by Ar Brandon last updated on 07/Sep/22

First term=common difference=common ratio= p 2^(nd) term of GP, p×p, equals 8^(th) term of AP, p+7p=8p ⇒p^2 =8p ⇒p^2 −8p=0 ⇒p=8 ⇒Sum of GP, S=((8(8^n −1))/7) ⇒S_5 =((8(8^5 −1))/7)=37448

$$\mathrm{First}\:\mathrm{term}=\mathrm{common}\:\mathrm{difference}=\mathrm{common}\:\mathrm{ratio}=\:{p} \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{term}\:\mathrm{of}\:{GP},\:{p}×{p},\:\mathrm{equals}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:{AP},\:{p}+\mathrm{7}{p}=\mathrm{8}{p} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\mathrm{8}{p}\:\Rightarrow{p}^{\mathrm{2}} −\mathrm{8}{p}=\mathrm{0}\:\Rightarrow{p}=\mathrm{8} \\ $$$$\Rightarrow\mathrm{Sum}\:\mathrm{of}\:{GP},\:{S}=\frac{\mathrm{8}\left(\mathrm{8}^{{n}} −\mathrm{1}\right)}{\mathrm{7}}\:\Rightarrow{S}_{\mathrm{5}} =\frac{\mathrm{8}\left(\mathrm{8}^{\mathrm{5}} −\mathrm{1}\right)}{\mathrm{7}}=\mathrm{37448}\: \\ $$

Commented by pete last updated on 10/Sep/22

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Sep/22

a=d=r AP: r,2r,...,7r,8r,9r,...,nr GP:r,r^2 ,r^3 ,...,r^n r^2 =8r⇒r=8=a S_n =((a(r^n −1))/(r−1)) S_5 =((r(r^5 −1))/(r−1))=((8(8^5 −1))/(8−1))=((8(8^5 −1))/7)

$${a}={d}={r} \\ $$$${AP}:\:{r},\mathrm{2}{r},...,\mathrm{7}{r},\mathrm{8}{r},\mathrm{9}{r},...,{nr} \\ $$$${GP}:{r},{r}^{\mathrm{2}} ,{r}^{\mathrm{3}} ,...,{r}^{{n}} \\ $$$${r}^{\mathrm{2}} =\mathrm{8}{r}\Rightarrow{r}=\mathrm{8}={a} \\ $$$${S}_{{n}} =\frac{{a}\left({r}^{{n}} −\mathrm{1}\right)}{{r}−\mathrm{1}} \\ $$$${S}_{\mathrm{5}} =\frac{{r}\left({r}^{\mathrm{5}} −\mathrm{1}\right)}{{r}−\mathrm{1}}=\frac{\mathrm{8}\left(\mathrm{8}^{\mathrm{5}} −\mathrm{1}\right)}{\mathrm{8}−\mathrm{1}}=\frac{\mathrm{8}\left(\mathrm{8}^{\mathrm{5}} −\mathrm{1}\right)}{\mathrm{7}} \\ $$

Commented by pete last updated on 10/Sep/22

Thanks sir

$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com