Note: 

\displaystyle S_{\infty} = \frac{a}{1-r}  

\displaystyle \textbf{Question 1: }\text{Find the sum of each of the following series to infinity:}
\displaystyle \text{(i) }1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\ldots
\displaystyle \text{(ii) }8+4\sqrt{2}+4+\ldots
\displaystyle \text{(iii) }\frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\ldots
\displaystyle \text{(iv) }10-9+8.1-7.29+\ldots
\displaystyle \text{(v) }\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\frac{1}{3^5}+\frac{1}{5^6}+\ldots
\displaystyle \textbf{Answer:}
\displaystyle \text{For an infinite G.P. with }|r|<1,\qquad S_{\infty}=\frac{a}{1-r}.
\displaystyle \text{(i) Given series: }1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\ldots
\displaystyle \text{Here, }a=1,\qquad r=\frac{-\frac{1}{3}}{1}=-\frac{1}{3}.
\displaystyle \text{Since }\left|-\frac{1}{3}\right|<1,\text{ the series is convergent.}
\displaystyle S_{\infty}=\frac{1}{1-\left(-\frac{1}{3}\right)}
\displaystyle =\frac{1}{1+\frac{1}{3}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}.
\displaystyle \therefore \text{The sum to infinity is }\frac{3}{4}.
\displaystyle \text{(ii) Given series: }8+4\sqrt{2}+4+\ldots
\displaystyle \text{Here, }a=8,\qquad r=\frac{4\sqrt{2}}{8}=\frac{1}{\sqrt{2}}.
\displaystyle \text{Since }\left|\frac{1}{\sqrt{2}}\right|<1,\text{ the series is convergent.}
\displaystyle S_{\infty}=\frac{8}{1-\frac{1}{\sqrt{2}}}
\displaystyle =\frac{8\sqrt{2}}{\sqrt{2}-1}
\displaystyle =\frac{8\sqrt{2}(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}
\displaystyle =8\sqrt{2}(\sqrt{2}+1)
\displaystyle =16+8\sqrt{2}=8(2+\sqrt{2}).
\displaystyle \therefore \text{The sum to infinity is }8(2+\sqrt{2}).
\displaystyle \text{(iii) Given series: }\frac{2}{5}+\frac{3}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\ldots
\displaystyle S_{\infty}=\left(\frac{2}{5}+\frac{2}{5^3}+\frac{2}{5^5}+\ldots\right)
\displaystyle \qquad+\left(\frac{3}{5^2}+\frac{3}{5^4}+\frac{3}{5^6}+\ldots\right)
\displaystyle =\frac{2}{5}\left(1+\frac{1}{5^2}+\frac{1}{5^4}+\ldots\right)
\displaystyle \qquad+\frac{3}{5^2}\left(1+\frac{1}{5^2}+\frac{1}{5^4}+\ldots\right)
\displaystyle =\left(\frac{2}{5}+\frac{3}{25}\right)\left(\frac{1}{1-\frac{1}{25}}\right)
\displaystyle =\frac{13}{25}\times\frac{25}{24}
\displaystyle =\frac{13}{24}.
\displaystyle \therefore \text{The sum to infinity is }\frac{13}{24}.
\displaystyle \text{(iv) Given series: }10-9+8.1-7.29+\ldots
\displaystyle \text{Here, }a=10,\qquad r=\frac{-9}{10}=-0.9.
\displaystyle \text{Since }|-0.9|<1,\text{ the series is convergent.}
\displaystyle S_{\infty}=\frac{10}{1-(-0.9)}
\displaystyle =\frac{10}{1.9}=\frac{100}{19}.
\displaystyle \therefore \text{The sum to infinity is }\frac{100}{19}.
\displaystyle \text{(v) Given series: }\frac{1}{3}+\frac{1}{5^2}+\frac{1}{3^3}+\frac{1}{5^4}+\ldots
\displaystyle S_{\infty}=\left(\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+\ldots\right)
\displaystyle \qquad+\left(\frac{1}{5^2}+\frac{1}{5^4}+\frac{1}{5^6}+\ldots\right)
\displaystyle =\frac{\frac{1}{3}}{1-\frac{1}{3^2}}+\frac{\frac{1}{5^2}}{1-\frac{1}{5^2}}
\displaystyle =\frac{1}{3}\left(\frac{9}{8}\right)+\frac{1}{25}\left(\frac{25}{24}\right)
\displaystyle =\frac{3}{8}+\frac{1}{24}
\displaystyle =\frac{9}{24}+\frac{1}{24}
\displaystyle =\frac{10}{24}=\frac{5}{12}.
\displaystyle \therefore \text{The sum to infinity is }\frac{5}{12}.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Prove that }9^{\frac{1}{3}}\cdot9^{\frac{1}{9}}\cdot9^{\frac{1}{27}}\cdot\ldots=3.
\displaystyle \textbf{Answer:}
\displaystyle \text{LHS}=9^{\frac{1}{3}}\cdot9^{\frac{1}{9}}\cdot9^{\frac{1}{27}}\cdot\ldots
\displaystyle =9^{\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots\right)}.
\displaystyle \text{The exponent is an infinite G.P. with first term }\frac{1}{3}\text{ and common ratio }\frac{1}{3}.
\displaystyle \therefore \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\ldots
\displaystyle =\frac{\frac{1}{3}}{1-\frac{1}{3}}
\displaystyle =\frac{1}{2}.
\displaystyle \therefore \text{LHS}=9^{\frac{1}{2}}=3=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that }2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots\infty=2.
\displaystyle \textbf{Answer:}
\displaystyle \text{LHS}=2^{\frac{1}{4}}\cdot4^{\frac{1}{8}}\cdot8^{\frac{1}{16}}\cdot16^{\frac{1}{32}}\cdots
\displaystyle =2^{\frac{1}{4}}\cdot(2^2)^{\frac{1}{8}}\cdot(2^3)^{\frac{1}{16}}\cdot(2^4)^{\frac{1}{32}}\cdots
\displaystyle =2^{\left(\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\cdots\right)}.
\displaystyle \text{Let }S=\frac{1}{4}+\frac{2}{8}+\frac{3}{16}+\frac{4}{32}+\cdots.
\displaystyle \therefore \frac{S}{2}=\frac{1}{8}+\frac{2}{16}+\frac{3}{32}+\frac{4}{64}+\cdots.
\displaystyle \text{Subtracting,}
\displaystyle S-\frac{S}{2}=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\cdots
\displaystyle \Rightarrow \frac{S}{2}=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{1}{2}.
\displaystyle \therefore S=1.
\displaystyle \therefore \text{LHS}=2^S=2^1=2=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }S_p\text{ denotes the sum of the series }1+r^p+r^{2p}+\ldots\text{ and }s_p
\displaystyle \text{denotes the sum of the series }1-r^p+r^{2p}-\ldots,\text{ prove that }S_p+s_p=2S_{2p}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Since the series are infinite G.P.s, we assume }|r|<1.
\displaystyle S_p=1+r^p+r^{2p}+\ldots=\frac{1}{1-r^p}.
\displaystyle s_p=1-r^p+r^{2p}-r^{3p}+\ldots=\frac{1}{1+r^p}.
\displaystyle \therefore S_p+s_p=\frac{1}{1-r^p}+\frac{1}{1+r^p}
\displaystyle =\frac{(1+r^p)+(1-r^p)}{(1-r^p)(1+r^p)}
\displaystyle =\frac{2}{1-r^{2p}}.
\displaystyle \text{But }S_{2p}=1+r^{2p}+r^{4p}+\ldots=\frac{1}{1-r^{2p}}.
\displaystyle \therefore S_p+s_p=2S_{2p}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the sum of the terms of an infinite decreasing G.P. in which all the terms}
\displaystyle \text{are positive, the first term is }4,\text{ and the difference between the third and fifth}
\displaystyle \text{terms is }\frac{32}{81}.
\displaystyle \textbf{Answer:}
\displaystyle \text{Given, }a=4.
\displaystyle \text{Let the common ratio be }r.
\displaystyle \text{Since the G.P. is decreasing and all its terms are positive, }0<r<1.
\displaystyle \text{Also, }a_3-a_5=\frac{32}{81}.
\displaystyle ar^2-ar^4=\frac{32}{81}
\displaystyle 4r^2-4r^4=\frac{32}{81}
\displaystyle r^2-r^4=\frac{8}{81}.
\displaystyle \text{Let }r^2=x.
\displaystyle x-x^2=\frac{8}{81}
\displaystyle 81x-81x^2=8
\displaystyle 81x^2-81x+8=0
\displaystyle (9x-8)(9x-1)=0.
\displaystyle \therefore x=\frac{8}{9}\qquad\text{or}\qquad x=\frac{1}{9}.
\displaystyle \text{Since }r>0,
\displaystyle r=\frac{2\sqrt{2}}{3}\qquad\text{or}\qquad r=\frac{1}{3}.
\displaystyle \text{For an infinite G.P., }S_{\infty}=\frac{a}{1-r}.
\displaystyle \text{When }r=\frac{2\sqrt{2}}{3},
\displaystyle S_{\infty}=\frac{4}{1-\frac{2\sqrt{2}}{3}}
\displaystyle =\frac{12}{3-2\sqrt{2}}
\displaystyle =\frac{12(3+2\sqrt{2})}{(3-2\sqrt{2})(3+2\sqrt{2})}
\displaystyle =12(3+2\sqrt{2})
\displaystyle =36+24\sqrt{2}.
\displaystyle \text{When }r=\frac{1}{3},
\displaystyle S_{\infty}=\frac{4}{1-\frac{1}{3}}
\displaystyle =\frac{4}{\frac{2}{3}}=6.
\displaystyle \therefore \text{The possible sums of the G.P. are }6\text{ and }36+24\sqrt{2}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Express the recurring decimal }0.\overline{125}\text{ as a rational number.}
\displaystyle \textbf{Answer:}
\displaystyle 0.\overline{125}=0.125125125\ldots
\displaystyle =0.125+0.000125+0.000000125+\ldots
\displaystyle =\frac{125}{10^3}+\frac{125}{10^6}+\frac{125}{10^9}+\ldots
\displaystyle =\frac{125}{10^3}\left(1+\frac{1}{10^3}+\frac{1}{10^6}+\ldots\right)
\displaystyle =\frac{125}{10^3}\left(\frac{1}{1-\frac{1}{10^3}}\right)
\displaystyle =\frac{125}{1000}\times\frac{1000}{999}
\displaystyle =\frac{125}{999}.
\displaystyle \therefore \text{The required rational number is }\frac{125}{999}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the rational number whose decimal expansion is }0.4\overline{23}.
\displaystyle \textbf{Answer:}
\displaystyle 0.4\overline{23}=0.4232323\ldots
\displaystyle =\frac{1}{10}\left(4.232323\ldots\right)
\displaystyle =\frac{1}{10}\left(4+0.23+0.0023+0.000023+\ldots\right)
\displaystyle =0.4+\frac{1}{10}\left(0.23+0.0023+0.000023+\ldots\right)
\displaystyle =0.4+\frac{1}{10}\left(\frac{23}{10^2}+\frac{23}{10^4}+\frac{23}{10^6}+\ldots\right)
\displaystyle =0.4+\frac{23}{10^3}\left(1+\frac{1}{10^2}+\frac{1}{10^4}+\ldots\right)
\displaystyle =0.4+\frac{23}{10^3}\left(\frac{1}{1-\frac{1}{10^2}}\right)
\displaystyle =0.4+\frac{23}{1000}\times\frac{100}{99}
\displaystyle =\frac{2}{5}+\frac{23}{990}
\displaystyle =\frac{396+23}{990}=\frac{419}{990}.
\displaystyle \therefore \text{The required rational number is }\frac{419}{990}.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the rational numbers having the following decimal expansions:}
\displaystyle \text{(i) }0.\overline{3}\qquad\text{(ii) }0.\overline{231}\qquad\text{(iii) }3.5\overline{2}\qquad\text{(iv) }0.6\overline{8}
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) }0.\overline{3}=0.33333\ldots
\displaystyle =0.3+0.03+0.003+\ldots
\displaystyle =\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+\ldots
\displaystyle =\frac{3}{10}\left(1+\frac{1}{10}+\frac{1}{10^2}+\ldots\right)
\displaystyle =\frac{3}{10}\left(\frac{1}{1-\frac{1}{10}}\right)
\displaystyle =\frac{3}{10}\times\frac{10}{9}=\frac{1}{3}.
\displaystyle \therefore \text{The required rational number is }\frac{1}{3}.
\displaystyle \text{(ii) }0.\overline{231}=0.231231231\ldots
\displaystyle =0.231+0.000231+0.000000231+\ldots
\displaystyle =\frac{231}{10^3}+\frac{231}{10^6}+\frac{231}{10^9}+\ldots
\displaystyle =\frac{231}{10^3}\left(1+\frac{1}{10^3}+\frac{1}{10^6}+\ldots\right)
\displaystyle =\frac{231}{10^3}\left(\frac{1}{1-\frac{1}{10^3}}\right)
\displaystyle =\frac{231}{10^3}\times\frac{10^3}{999}=\frac{231}{999}=\frac{77}{333}.
\displaystyle \therefore \text{The required rational number is }\frac{77}{333}.
\displaystyle \text{(iii) }3.5\overline{2}=3.522222\ldots
\displaystyle =3.5+0.02+0.002+0.0002+\ldots
\displaystyle =3.5+\frac{2}{100}+\frac{2}{10^3}+\frac{2}{10^4}+\ldots
\displaystyle =3.5+\frac{2}{100}\left(1+\frac{1}{10}+\frac{1}{10^2}+\ldots\right)
\displaystyle =3.5+\frac{2}{100}\left(\frac{1}{1-\frac{1}{10}}\right)
\displaystyle =\frac{35}{10}+\frac{2}{100}\times\frac{10}{9}
\displaystyle =\frac{35}{10}+\frac{2}{90}=\frac{317}{90}.
\displaystyle \therefore \text{The required rational number is }\frac{317}{90}.
\displaystyle \text{(iv) }0.6\overline{8}=0.688888\ldots
\displaystyle =0.6+0.08+0.008+0.0008+\ldots
\displaystyle =0.6+\frac{8}{100}+\frac{8}{10^3}+\frac{8}{10^4}+\ldots
\displaystyle =0.6+\frac{8}{100}\left(1+\frac{1}{10}+\frac{1}{10^2}+\ldots\right)
\displaystyle =0.6+\frac{8}{100}\left(\frac{1}{1-\frac{1}{10}}\right)
\displaystyle =\frac{6}{10}+\frac{8}{100}\times\frac{10}{9}
\displaystyle =\frac{6}{10}+\frac{8}{90}=\frac{62}{90}=\frac{31}{45}.
\displaystyle \therefore \text{The required rational number is }\frac{31}{45}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{One side of an equilateral triangle is }18\text{ cm. The mid-points of its sides}
\displaystyle \text{are joined to form another triangle, whose mid-points, in turn, are joined to form}
\displaystyle \text{another triangle. The process is continued indefinitely. Find the sum of the}
\displaystyle \text{(i) perimeters and (ii) areas of all the triangles.}
\displaystyle \textbf{Answer:}
\displaystyle \text{The side of the first triangle is }18\text{ cm.}
\displaystyle \text{The side of the second triangle is }\frac{1}{2}\times18=9\text{ cm.}
\displaystyle \text{The side of the third triangle is }\frac{1}{2}\times9=\frac{9}{2}\text{ cm.}
\displaystyle \therefore \text{The side lengths form the infinite G.P. }18,9,\frac{9}{2},\ldots
\displaystyle \text{(i) Sum of the perimeters of all the triangles}
\displaystyle =3\left(18+9+\frac{9}{2}+\ldots\right)
\displaystyle =3\left(\frac{18}{1-\frac{1}{2}}\right)
\displaystyle =3\times36=108\text{ cm}.
\displaystyle \therefore \text{The sum of the perimeters of all the triangles is }108\text{ cm}.
\displaystyle \text{(ii) Area of an equilateral triangle of side }a\text{ is }\frac{\sqrt{3}}{4}a^2.
\displaystyle \therefore \text{Sum of the areas of all the triangles}
\displaystyle =\frac{\sqrt{3}}{4}\left[18^2+9^2+\left(\frac{9}{2}\right)^2+\ldots\right]
\displaystyle =\frac{\sqrt{3}}{4}\left(\frac{18^2}{1-\frac{1}{4}}\right)
\displaystyle =\frac{\sqrt{3}}{4}\left(\frac{324}{\frac{3}{4}}\right)
\displaystyle =\frac{\sqrt{3}}{4}\times432
\displaystyle =108\sqrt{3}\text{ cm}^2.
\displaystyle \therefore \text{The sum of the areas of all the triangles is }108\sqrt{3}\text{ cm}^2.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find an infinite G.P. whose first term is }1\text{ and each term is the sum of}
\displaystyle \text{all the terms which follow it.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the common ratio be }r.
\displaystyle \therefore \text{The G.P. is }1,r,r^2,r^3,\ldots
\displaystyle \text{The sum of all the terms following the first term is}
\displaystyle r+r^2+r^3+\ldots=\frac{r}{1-r}.
\displaystyle \text{Given, the first term is equal to the sum of all the terms following it.}
\displaystyle 1=\frac{r}{1-r}
\displaystyle 1-r=r
\displaystyle 2r=1
\displaystyle \therefore r=\frac{1}{2}.
\displaystyle \therefore \text{The required G.P. is }1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\ldots
\displaystyle \text{Also, for any term }r^k,
\displaystyle r^{k+1}+r^{k+2}+\ldots=\frac{r^{k+1}}{1-r}=r^k\text{ when }r=\frac{1}{2}.
\displaystyle \therefore \text{Each term is the sum of all the terms which follow it.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{The sum of the first two terms of an infinite G.P. is }5,\text{ and each term}
\displaystyle \text{is three times the sum of all the succeeding terms. Find the G.P.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle \therefore \text{The G.P. is }a,ar,ar^2,ar^3,\ldots
\displaystyle \text{Given, the sum of the first two terms is }5.
\displaystyle a+ar=5
\displaystyle a(1+r)=5.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{The sum of all the terms succeeding the first term is}
\displaystyle ar+ar^2+ar^3+\ldots=\frac{ar}{1-r}.
\displaystyle \text{Since each term is three times the sum of all the succeeding terms,}
\displaystyle a=3\left(\frac{ar}{1-r}\right)
\displaystyle 1=\frac{3r}{1-r}
\displaystyle 1-r=3r
\displaystyle 4r=1
\displaystyle \therefore r=\frac{1}{4}.
\displaystyle \text{Substituting }r=\frac{1}{4}\text{ in (i),}
\displaystyle a\left(1+\frac{1}{4}\right)=5
\displaystyle a\left(\frac{5}{4}\right)=5
\displaystyle \therefore a=4.
\displaystyle \therefore \text{The required G.P. is }4,1,\frac{1}{4},\frac{1}{16},\ldots
\displaystyle \text{Indeed, for any term }ar^k,
\displaystyle \text{the sum of all the succeeding terms}=\frac{ar^{k+1}}{1-r}.
\displaystyle 3\left(\frac{ar^{k+1}}{1-r}\right)=ar^k\text{ when }r=\frac{1}{4}.
\displaystyle \therefore \text{The given condition is satisfied by every term.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Show that in an infinite G.P. with common ratio }r,\text{ where }0<|r|<1,
\displaystyle \text{each term bears a constant ratio to the sum of all the terms that follow it.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the infinite G.P. be }a,ar,ar^2,ar^3,\ldots
\displaystyle \text{Its }n^{\mathrm{th}}\text{ term is }a_n=ar^{n-1}.
\displaystyle \text{Let }T_n\text{ denote the sum of all the terms following }a_n.
\displaystyle \therefore T_n=ar^n+ar^{n+1}+ar^{n+2}+\ldots
\displaystyle =\frac{ar^n}{1-r}.
\displaystyle \therefore \frac{a_n}{T_n}=\frac{ar^{n-1}}{\frac{ar^n}{1-r}}
\displaystyle =\frac{1-r}{r}.
\displaystyle \text{Since }\frac{1-r}{r}\text{ is independent of }n,\text{ the ratio is constant.}
\displaystyle \therefore \text{Each term bears the constant ratio }\frac{1-r}{r}\text{ to the sum of all}
\displaystyle \text{the terms that follow it. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }S\text{ denotes the sum of an infinite G.P. and }S_1\text{ denotes the sum}
\displaystyle \text{of the squares of its terms, prove that the first term and common ratio are}
\displaystyle \frac{2SS_1}{S^2+S_1}\text{ and }\frac{S^2-S_1}{S^2+S_1}\text{ respectively.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the first term be }a\neq0\text{ and the common ratio be }r,\text{ where }|r|<1.
\displaystyle \text{Given, }S=a+ar+ar^2+\ldots
\displaystyle \therefore S=\frac{a}{1-r}.\qquad\ldots\ldots\text{(i)}
\displaystyle \text{Also, }S_1=a^2+a^2r^2+a^2r^4+\ldots
\displaystyle \therefore S_1=\frac{a^2}{1-r^2}.\qquad\ldots\ldots\text{(ii)}
\displaystyle \text{From (i),}
\displaystyle S^2=\frac{a^2}{(1-r)^2}.
\displaystyle \text{Using (ii), }a^2=S_1(1-r^2).
\displaystyle \therefore S^2=\frac{S_1(1-r^2)}{(1-r)^2}
\displaystyle =S_1\left(\frac{(1-r)(1+r)}{(1-r)^2}\right)
\displaystyle =S_1\left(\frac{1+r}{1-r}\right).
\displaystyle (1-r)S^2=(1+r)S_1
\displaystyle S^2-rS^2=S_1+rS_1
\displaystyle S^2-S_1=r(S^2+S_1)
\displaystyle \therefore r=\frac{S^2-S_1}{S^2+S_1}.
\displaystyle \text{Substituting this value of }r\text{ in (i),}
\displaystyle S=\frac{a}{1-\frac{S^2-S_1}{S^2+S_1}}
\displaystyle =a\left[\frac{S^2+S_1}{(S^2+S_1)-(S^2-S_1)}\right]
\displaystyle =a\left(\frac{S^2+S_1}{2S_1}\right).
\displaystyle \therefore a=\frac{2SS_1}{S^2+S_1}.
\displaystyle \therefore \text{The first term is }\frac{2SS_1}{S^2+S_1}\text{ and the common ratio is}
\displaystyle \frac{S^2-S_1}{S^2+S_1}.\text{ Hence proved.}
\displaystyle \\

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