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In-a-geometric-sequence-of-real-numbers-the-sum-of-the-first-two-terms-is-7-and-the-sum-of-the-first-six-terms-is-91-The-sum-of-the-first-four-terms-is-

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Question Number 111542 by Aina Samuel Temidayo last updated on 04/Sep/20
In a geometric sequence of real numbers, the sum of the first two terms is 7 and the sum of the first six terms is 91. The sum of the first four terms is.
$$\mathrm{In}\:\mathrm{a}\:\mathrm{geometric}\:\mathrm{sequence}\:\mathrm{of}\:\mathrm{real} \\ $$$$\mathrm{numbers},\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{two}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{7}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{six}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{91}.\:\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{four}\:\mathrm{terms}\:\mathrm{is}. \\ $$$$ \\ $$
Commented by bemath last updated on 04/Sep/20
what the meaning sum of two two terms?
$${what}\:{the}\:{meaning}\:{sum}\:{of}\:{two}\:{two}\:{terms}? \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
I meant the sum of the first two terms.
$$\mathrm{I}\:\mathrm{meant}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{two}\:\mathrm{terms}. \\ $$
Answered by Lordose last updated on 04/Sep/20
S_2 =((a(1−r^2 ))/(1−r))=a(1+r)=7⇒(1) S_6 =((a(1−r^6 ))/(1−r))=a(1+r)(r^4 +r^2 +1)=91⇒(2) (((2))/((1)))=r^4 +r^2 +1=13 set r^2 =x x^2 +x−12=0 x^2 +4x−3x−12=0 x(x+4)−3(x+4)=0 x=3 or x=−4 r=(√3) or r=2i Since r can′t be complex, r=(√3) Sub in (1) a(1+(√3))=7 a= (7/(1+(√3))) S_4 =((a(1−r^4 ))/(1−r))= ((7(1−((√3))^4 ))/((1+(√3))(1−(√3)))) S_4 =((7(−8))/(−2))=28 ★LorD OsE
$$\boldsymbol{\mathrm{S}}_{\mathrm{2}} =\frac{\boldsymbol{\mathrm{a}}\left(\mathrm{1}−\boldsymbol{\mathrm{r}}^{\mathrm{2}} \right)}{\mathrm{1}−\boldsymbol{\mathrm{r}}}=\boldsymbol{\mathrm{a}}\left(\mathrm{1}+\boldsymbol{\mathrm{r}}\right)=\mathrm{7}\Rightarrow\left(\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{6}} =\frac{\boldsymbol{\mathrm{a}}\left(\mathrm{1}−\boldsymbol{\mathrm{r}}^{\mathrm{6}} \right)}{\mathrm{1}−\boldsymbol{\mathrm{r}}}=\boldsymbol{\mathrm{a}}\left(\mathrm{1}+\boldsymbol{\mathrm{r}}\right)\left(\boldsymbol{\mathrm{r}}^{\mathrm{4}} +\boldsymbol{\mathrm{r}}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{91}\Rightarrow\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{2}\right)}{\left(\mathrm{1}\right)}=\boldsymbol{\mathrm{r}}^{\mathrm{4}} +\boldsymbol{\mathrm{r}}^{\mathrm{2}} +\mathrm{1}=\mathrm{13} \\ $$$$\boldsymbol{\mathrm{set}}\:\boldsymbol{\mathrm{r}}^{\mathrm{2}} =\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}−\mathrm{12}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{\mathrm{x}}−\mathrm{3}\boldsymbol{\mathrm{x}}−\mathrm{12}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)−\mathrm{3}\left(\boldsymbol{\mathrm{x}}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{x}}=\mathrm{3}\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{x}}=−\mathrm{4} \\ $$$$\boldsymbol{\mathrm{r}}=\sqrt{\mathrm{3}}\:\:\boldsymbol{\mathrm{or}}\:\boldsymbol{\mathrm{r}}=\mathrm{2}\boldsymbol{\mathrm{i}} \\ $$$$\boldsymbol{\mathrm{Since}}\:\boldsymbol{\mathrm{r}}\:\boldsymbol{\mathrm{can}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{complex}},\:\boldsymbol{\mathrm{r}}=\sqrt{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{Sub}}\:\boldsymbol{\mathrm{in}}\:\left(\mathrm{1}\right) \\ $$$$\boldsymbol{\mathrm{a}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)=\mathrm{7} \\ $$$$\boldsymbol{\mathrm{a}}=\:\frac{\mathrm{7}}{\mathrm{1}+\sqrt{\mathrm{3}}} \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{4}} =\frac{\boldsymbol{\mathrm{a}}\left(\mathrm{1}−\boldsymbol{\mathrm{r}}^{\mathrm{4}} \right)}{\mathrm{1}−\boldsymbol{\mathrm{r}}}=\:\frac{\mathrm{7}\left(\mathrm{1}−\left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \right)}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)} \\ $$$$\boldsymbol{\mathrm{S}}_{\mathrm{4}} =\frac{\mathrm{7}\left(−\mathrm{8}\right)}{−\mathrm{2}}=\mathrm{28} \\ $$$$\bigstar\boldsymbol{\mathrm{LorD}}\:\boldsymbol{\mathrm{OsE}} \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
Thanks.
$$\mathrm{Thanks}. \\ $$
Answered by Rasheed.Sindhi last updated on 04/Sep/20
a+ar=7 a+ar+ar^2 +ar^3 +ar^4 +ar^5 =91 a+ar+ar^2 +ar^3 =? a=(7/(1+r)) a+ar+r^2 (a+ar+ar^2 +ar^3 )=91 a+ar+ar^2 +ar^3 =((91−(a+ar))/r^2 ) =((91−7)/r^2 )=((84)/r^2 ) a+ar+ar^2 +ar^3 +ar^4 +ar^5 =91 a(1+r+r^2 +...+r^5 )=91 ((7/(1+r)))(1+r+r^2 +...+r^5 )=91 ((1/(1+r)))(1+r+r^2 +...+r^5 )=13 ((1−r^6 )/(1−r^2 ))=13 (((1−r^2 )(1+r^2 +r^4 ))/(1−r^2 ))=13 r^4 +r^2 −12=0 (r^2 +4)(r^2 −3)=0 r^2 =−4_(r is not real) ∣ r^2 =3 Sum of first four terms=((84)/r^2 ) = ((84)/3)=28 (corrected)
$${a}+{ar}=\mathrm{7} \\ $$$${a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} +{ar}^{\mathrm{4}} +{ar}^{\mathrm{5}} =\mathrm{91} \\ $$$${a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} =? \\ $$$${a}=\frac{\mathrm{7}}{\mathrm{1}+{r}} \\ $$$${a}+{ar}+{r}^{\mathrm{2}} \left({a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} \right)=\mathrm{91} \\ $$$${a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} =\frac{\mathrm{91}−\left({a}+{ar}\right)}{{r}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{91}−\mathrm{7}}{{r}^{\mathrm{2}} }=\frac{\mathrm{84}}{{r}^{\mathrm{2}} } \\ $$$${a}+{ar}+{ar}^{\mathrm{2}} +{ar}^{\mathrm{3}} +{ar}^{\mathrm{4}} +{ar}^{\mathrm{5}} =\mathrm{91} \\ $$$${a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +…+{r}^{\mathrm{5}} \right)=\mathrm{91} \\ $$$$\left(\frac{\mathrm{7}}{\mathrm{1}+{r}}\right)\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +…+{r}^{\mathrm{5}} \right)=\mathrm{91} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{1}+{r}}\right)\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} +…+{r}^{\mathrm{5}} \right)=\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}−{r}^{\mathrm{6}} }{\mathrm{1}−{r}^{\mathrm{2}} }=\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\frac{\left(\mathrm{1}−{r}^{\mathrm{2}} \right)\left(\mathrm{1}+{r}^{\mathrm{2}} +{r}^{\mathrm{4}} \right)}{\mathrm{1}−{r}^{\mathrm{2}} }=\mathrm{13} \\ $$$$\:\:\:\:\:\:\:{r}^{\mathrm{4}} +{r}^{\mathrm{2}} −\mathrm{12}=\mathrm{0} \\ $$$$\:\:\:\:\left({r}^{\mathrm{2}} +\mathrm{4}\right)\left({r}^{\mathrm{2}} −\mathrm{3}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\underset{{r}\:{is}\:{not}\:{real}} {\:{r}^{\mathrm{2}} =−\mathrm{4}}\:\mid\:{r}^{\mathrm{2}} =\mathrm{3} \\ $$$$ \\ $$$${Sum}\:{of}\:{first}\:{four}\:{terms}=\frac{\mathrm{84}}{{r}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{84}}{\mathrm{3}}=\mathrm{28} \\ $$$$\left({corrected}\right) \\ $$
Commented by Aina Samuel Temidayo last updated on 04/Sep/20
Seen.
$$\mathrm{Seen}. \\ $$

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