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the-third-sixth-and-seventh-terms-of-a-geometric-progression-whose-common-ratio-is-neither-0-nor-1-are-in-arithmetic-progression-prove-that-the-sum-of-the-first-three-terms-is-equal-to-fourth-




Question Number 103780 by bobhans last updated on 17/Jul/20
the third, sixth and seventh terms of a  geometric progression (whose common  ratio is neither 0 nor 1 ) are in  arithmetic progression . prove that the  sum of the first three terms is equal to  fourth .
$${the}\:{third},\:{sixth}\:{and}\:{seventh}\:{terms}\:{of}\:{a} \\ $$$${geometric}\:{progression}\:\left({whose}\:{common}\right. \\ $$$$\left.{ratio}\:{is}\:{neither}\:\mathrm{0}\:{nor}\:\mathrm{1}\:\right)\:{are}\:{in} \\ $$$${arithmetic}\:{progression}\:.\:{prove}\:{that}\:{the} \\ $$$${sum}\:{of}\:{the}\:{first}\:{three}\:{terms}\:{is}\:{equal}\:{to} \\ $$$${fourth}\:. \\ $$
Answered by bemath last updated on 17/Jul/20
AP : ar^2  , ar^5 , ar^6  ⇒2ar^5  = ar^2 +ar^6   2r^5  = r^2 +r^6  ⇒ 2r^3  = 1+r^4   r^4 −2r^3 +1 = 0 ; (r−1)(r^3 −r^2 −r−1)=0  r^3 = r^2 +r + 1  now we have T_4 =ar^3   sum of the first three terms is   a+ar+ar^2  = a(1+r+r^2 )   = a(r^3 ) = T_4  . proved ■
$${AP}\::\:{ar}^{\mathrm{2}} \:,\:{ar}^{\mathrm{5}} ,\:{ar}^{\mathrm{6}} \:\Rightarrow\mathrm{2}{ar}^{\mathrm{5}} \:=\:{ar}^{\mathrm{2}} +{ar}^{\mathrm{6}} \\ $$$$\mathrm{2}{r}^{\mathrm{5}} \:=\:{r}^{\mathrm{2}} +{r}^{\mathrm{6}} \:\Rightarrow\:\mathrm{2}{r}^{\mathrm{3}} \:=\:\mathrm{1}+{r}^{\mathrm{4}} \\ $$$${r}^{\mathrm{4}} −\mathrm{2}{r}^{\mathrm{3}} +\mathrm{1}\:=\:\mathrm{0}\:;\:\left({r}−\mathrm{1}\right)\left({r}^{\mathrm{3}} −{r}^{\mathrm{2}} −{r}−\mathrm{1}\right)=\mathrm{0} \\ $$$${r}^{\mathrm{3}} =\:{r}^{\mathrm{2}} +{r}\:+\:\mathrm{1} \\ $$$${now}\:{we}\:{have}\:{T}_{\mathrm{4}} ={ar}^{\mathrm{3}} \\ $$$${sum}\:{of}\:{the}\:{first}\:{three}\:{terms}\:{is}\: \\ $$$${a}+{ar}+{ar}^{\mathrm{2}} \:=\:{a}\left(\mathrm{1}+{r}+{r}^{\mathrm{2}} \right)\: \\ $$$$=\:{a}\left({r}^{\mathrm{3}} \right)\:=\:{T}_{\mathrm{4}} \:.\:{proved}\:\blacksquare \\ $$

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