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Question Number 29645 by gyugfeet last updated on 11/Feb/18

if the sum of first 5 terms of a G.P. is 155, sum of last 5 terms is 39680,first term is 5 and last term is 20480. find the number of terms of the sequence.

$${if}\:{the}\:{sum}\:{of}\:{first}\:\mathrm{5}\:{terms}\:{of}\:\:{a}\:{G}.{P}.\:{is}\:\mathrm{155},\:{sum}\:{of}\:{last}\:\mathrm{5}\:{terms}\:{is}\:\mathrm{39680},{first}\:{term}\:{is}\:\mathrm{5}\:{and}\:{last}\:{term}\:\:{is}\:\mathrm{20480}.\:{find}\:{the}\:{number}\:{of}\:{terms}\:{of}\:{the}\:{sequence}. \\ $$

Answered by Rasheed.Sindhi last updated on 11/Feb/18

Let the common ratio of the GP is r Sum of first five terms: 5+5r+5r^2 +5r^3 +5r^4 =155 Or 1+r+r^2 +r^3 +r^4 =31...............A Sum of last five terms(in reverse order): 20480+((20480)/r)+((20480)/r^2 )+((20480)/r^3 )+((20480)/r^4 )=39680 20480(1+(1/r)+(1/r^2 )+(1/r^3 )+(1/r^4 ))=39680 1+(1/r)+(1/r^2 )+(1/r^3 )+(1/r^4 )=((39680)/(20480)) (1/r^4 )(r^4 +r^3 +r^2 +r+1)=((39680)/(20480)) But from A, r^4 +r^3 +r^2 +r+1=31 (1/r^4 )(31)=((39680)/(20480)) r^4 =((20480×31)/(39680))=16 r=±2 Let the number of terms of GP is n Last term=ar^(n−1) a=5 (First term) , r=±2 , Last term=39680 5(±2)^(n−1) =20480 (±2)^(n−1) =4096 (±2)^(n−1) =(±2)^(12) n−1=12 n=13 The GP contains 13 terms

$$\mathrm{Let}\:\mathrm{the}\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{GP}\:\mathrm{is}\:\mathrm{r} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{five}\:\mathrm{terms}: \\ $$$$\mathrm{5}+\mathrm{5r}+\mathrm{5r}^{\mathrm{2}} +\mathrm{5r}^{\mathrm{3}} +\mathrm{5r}^{\mathrm{4}} =\mathrm{155} \\ $$$$\:\mathrm{Or}\:\:\mathrm{1}+\mathrm{r}+\mathrm{r}^{\mathrm{2}} +\mathrm{r}^{\mathrm{3}} +\mathrm{r}^{\mathrm{4}} =\mathrm{31}...............\mathrm{A} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{last}\:\mathrm{five}\:\mathrm{terms}\left(\mathrm{in}\:\mathrm{reverse}\:\mathrm{order}\right): \\ $$$$\mathrm{20480}+\frac{\mathrm{20480}}{\mathrm{r}}+\frac{\mathrm{20480}}{\mathrm{r}^{\mathrm{2}} }+\frac{\mathrm{20480}}{\mathrm{r}^{\mathrm{3}} }+\frac{\mathrm{20480}}{\mathrm{r}^{\mathrm{4}} }=\mathrm{39680} \\ $$$$\mathrm{20480}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{r}}+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{4}} }\right)=\mathrm{39680} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{r}}+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{4}} }=\frac{\mathrm{39680}}{\mathrm{20480}} \\ $$$$\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{4}} }\left(\mathrm{r}^{\mathrm{4}} +\mathrm{r}^{\mathrm{3}} +\mathrm{r}^{\mathrm{2}} +\mathrm{r}+\mathrm{1}\right)=\frac{\mathrm{39680}}{\mathrm{20480}} \\ $$$$\mathrm{But}\:\mathrm{from}\:\mathrm{A},\:\mathrm{r}^{\mathrm{4}} +\mathrm{r}^{\mathrm{3}} +\mathrm{r}^{\mathrm{2}} +\mathrm{r}+\mathrm{1}=\mathrm{31}\:\: \\ $$$$\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{4}} }\left(\mathrm{31}\right)=\frac{\mathrm{39680}}{\mathrm{20480}}\:\: \\ $$$$\mathrm{r}^{\mathrm{4}} =\frac{\mathrm{20480}×\mathrm{31}}{\mathrm{39680}}=\mathrm{16} \\ $$$$\mathrm{r}=\pm\mathrm{2} \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{GP}\:\mathrm{is}\:\mathrm{n} \\ $$$$\mathrm{Last}\:\mathrm{term}={ar}^{{n}−\mathrm{1}} \\ $$$${a}=\mathrm{5}\:\left(\mathrm{First}\:\mathrm{term}\right)\:,\:\mathrm{r}=\pm\mathrm{2}\:,\:\mathrm{Last}\:\mathrm{term}=\mathrm{39680} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{5}\left(\pm\mathrm{2}\right)^{\mathrm{n}−\mathrm{1}} =\mathrm{20480} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\pm\mathrm{2}\right)^{\mathrm{n}−\mathrm{1}} =\mathrm{4096} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left(\pm\mathrm{2}\right)^{\mathrm{n}−\mathrm{1}} =\left(\pm\mathrm{2}\right)^{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}−\mathrm{1}=\mathrm{12} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}=\mathrm{13} \\ $$$$\mathrm{The}\:\mathrm{GP}\:\mathrm{contains}\:\mathrm{13}\:\mathrm{terms} \\ $$

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