Question-179281

Question Number 179281 by cherokeesay last updated on 27/Oct/22

Commented by a.lgnaoui last updated on 28/Oct/22

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Answered by ARUNG_Brandon_MBU last updated on 28/Oct/22

For X: X_6 =(4a)r^5 =256 For Y: Y_5 =(3a)r^4 =48 (X_6 /Y_5 ) ⇒(4/3)r=((16)/3) ⇒r=4 ⇒4a=((256)/r^5 )=(1/4) ⇒a=(1/(16)) (i) X_1 =(1/4) (ii) Σ_(i=1) ^4 X_i =(1/4)∙((4^4 −1)/(4−1))=((2^8 −1)/(12))=((255)/(12))

$$\mathrm{For}\:\mathrm{X}:\:{X}_{\mathrm{6}} =\left(\mathrm{4}{a}\right){r}^{\mathrm{5}} =\mathrm{256} \\ $$$$\mathrm{For}\:\mathrm{Y}:\:{Y}_{\mathrm{5}} =\left(\mathrm{3}{a}\right){r}^{\mathrm{4}} =\mathrm{48} \\ $$$$\frac{\mathrm{X}_{\mathrm{6}} }{\mathrm{Y}_{\mathrm{5}} }\:\Rightarrow\frac{\mathrm{4}}{\mathrm{3}}{r}=\frac{\mathrm{16}}{\mathrm{3}}\:\Rightarrow{r}=\mathrm{4}\: \\ $$$$\Rightarrow\mathrm{4}{a}=\frac{\mathrm{256}}{{r}^{\mathrm{5}} }=\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\left({i}\right)\:\mathrm{X}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({ii}\right)\:\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\sum}}{X}_{{i}} =\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\mathrm{4}^{\mathrm{4}} −\mathrm{1}}{\mathrm{4}−\mathrm{1}}=\frac{\mathrm{2}^{\mathrm{8}} −\mathrm{1}}{\mathrm{12}}=\frac{\mathrm{255}}{\mathrm{12}} \\ $$

Commented by cherokeesay last updated on 28/Oct/22