Question

Consider two different infinite geometric progressions with their sums $S _1$ and $S _2$ as
$S _1= a + ar + ar ^2+ ar ^3+\ldots \ldots \ldots . . \infty$
$S _2= b + bR + bR ^2+ bR ^3+\ldots \ldots \ldots \ldots\infty$
If $S _1= S _2=1, ar = bR$ and $ar ^2=\frac{1}{8}$ 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
$a\, ar\, ar^2 .......... | r | < 1$
$b \,br \,br^2 .......... | R | < 1$
now $\frac{ a }{1- r }=1 ; \frac{ b }{1- R }=1$
$a=1-r$.....(1) and$b =1- R\ldots(2) \Rightarrow a+r=b+Q$
also $ ar $=b R$....(3) and $\operatorname{ar}^2=\frac{1}{8}$
(1) - (2) $a - b = R - r$
$a + r = b + R$
From (1), (2) and (3), $ar = bR$
$(1 - r) r = (1 - R)R$ (substituting a and b for (1) and (2) in (3))
$\therefore (r - R)(r + R) = r - R$
$\therefore r+R=1 ($ as $r \neq R)$
(Note $:$ if $r=R$ then $a = b \Rightarrow$ G.P. will become identical) hence $r + R =1 $....(5) hence $(1)+(2) \Rightarrow a+b=2-(r+R) \Rightarrow a+b=1$ now, from (4)
$8 ar ^2=1 $
$8(1-r) r ^2=1$
$8 r ^2-8 r ^3-1=0 \Rightarrow 8 r^3-8 r ^2+1=0 $
$(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) } $
$\therefore 4 r^2-2 r-1=0$
$r =\frac{2 \pm \sqrt{20}}{8}=\frac{1+\sqrt{5}}{4}$ or $\frac{1-\sqrt{5}}{4}$
if $r =\frac{1-\sqrt{5}}{4}$ then $R =1-\frac{1-\sqrt{5}}{4}=\frac{3+\sqrt{5}}{4}>1$
hence $r=\frac{1-\sqrt{5}}{4}$ is rejected
$\therefore r =\frac{1+\sqrt{5}}{4}$ Ans.
and
$R =1-\frac{1+\sqrt{5}}{4}=\frac{3-\sqrt{5}}{4} $
$a =1- r =\frac{3-\sqrt{5}}{4} ; b =\frac{1+\sqrt{5}}{4} ; a + b =1$