Question

Consider two different infinite geometric progressions with their sums $S_1$ and $S_2$ as
$S_1=a+ar+a{r}^2+a{r}^3+\dots \dots \dots ..\infty$
$S_2=b+bR+b{R}^2+b{R}^3+\dots \dots \dots \dots \infty$
If $S_1=S_2=1,ar=bR$ and $a{r}^2=\frac{1}{8}$ then answer the following:
Common ratio of the second G.P. is

• A. $\frac{3+\sqrt{5}}{4}$
• B. $\frac{3-\sqrt{5}}{4}$
• C. $\frac{1}{2}$
• D. none

Answer 1

Answer: (B)

Let the two GP's are
$aara{r}^2..........|r|<1$
$bbrb{r}^2..........|R|<1$
now $\frac{a}{1-r}=1;\frac{b}{1-R}=1$
$a=1-r$.....(1) and$b=1-R\dots (2)\Rightarrow a+r=b+Q$
also $ar$=b R$....(3)and$\operatorname{ar}^2=\frac{1}{8}$<br/>(1)-(2)$a - b = R - r$<br/>$a + r = b + R$<br/>From(1),(2)and(3),$ar = bR$<br/>$(1 - r) r = (1 - R)R$(substitutingaandbfor(1)and(2)in(3))<br/>$\therefore (r - R)(r + R) = r - R$<br/>$\therefore r+R=1 ($as$r \neq R)$<br/>(Note$:$if$r=R$then$a = b \Rightarrow$G.P.willbecomeidentical)hence$r + R =1 $....(5)hence$(1)+(2) \Rightarrow a+b=2-(r+R) \Rightarrow a+b=1$now,from(4)<br/>$8 ar ^2=1 $<br/>$8(1-r) r ^2=1$<br/>$8 r ^2-8 r ^3-1=0 \Rightarrow 8 r^3-8 r ^2+1=0 $<br/>$(2 r -1)\left(4 r ^2-2 r -1\right)=0 (\text { as } r \neq 1 / 2 \text { as in this case } R \text { will also be } 1 / 2 \text { which is not possible) } $<br/>$\therefore 4 r^2-2 r-1=0$<br/>$r =\frac{2 \pm \sqrt{20}}{8}=\frac{1+\sqrt{5}}{4}$or$\frac{1-\sqrt{5}}{4}$<br/>if$r =\frac{1-\sqrt{5}}{4}$then$R =1-\frac{1-\sqrt{5}}{4}=\frac{3+\sqrt{5}}{4}>1$<br/>hence$r=\frac{1-\sqrt{5}}{4}$isrejected<br/>$\therefore r =\frac{1+\sqrt{5}}{4}$Ans.<br/>and<br/>$R =1-\frac{1+\sqrt{5}}{4}=\frac{3-\sqrt{5}}{4} $<br/>$a =1- r =\frac{3-\sqrt{5}}{4} ; b =\frac{1+\sqrt{5}}{4} ; a + b =1$