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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-th

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

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