Class 11: Geometric Progression

$\displaystyle \textbf{Question 1: }\text{Show that each one of the following progressions is a G.P. Also, find the}$
$\displaystyle \text{common ratio in each case:}$
$\displaystyle \text{i) }4,-2,1,-\frac{1}{2},\ldots\qquad \text{ii) }-\frac{2}{3},-6,-54,\ldots$
$\displaystyle \text{iii) }a,\frac{3a^2}{4},\frac{9a^3}{16},\ldots\qquad \text{iv) }\frac{1}{2},\frac{1}{3},\frac{2}{9},\frac{4}{27},\ldots$
$\displaystyle \text{Answer:}$

$\displaystyle \text{i) Given progression: }4,-2,1,-\frac{1}{2},\ldots$
$\displaystyle \frac{-2}{4}=-\frac{1}{2},\qquad \frac{1}{-2}=-\frac{1}{2},\qquad \frac{-1/2}{1}=-\frac{1}{2}$
$\displaystyle \therefore \frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=-\frac{1}{2}$
$\displaystyle \text{Since the ratio of each term to its preceding term is constant, the progression is a G.P.}$
$\displaystyle \therefore \text{The first term is }4\text{ and the common ratio is }-\frac{1}{2}.$

$\displaystyle \text{ii) Given progression: }-\frac{2}{3},-6,-54,\ldots$
$\displaystyle \frac{-6}{-2/3}=9,\qquad \frac{-54}{-6}=9$
$\displaystyle \therefore \frac{a_2}{a_1}=\frac{a_3}{a_2}=9$
$\displaystyle \text{Since the ratio of each term to its preceding term is constant, the progression is a G.P.}$
$\displaystyle \therefore \text{The first term is }-\frac{2}{3}\text{ and the common ratio is }9.$

$\displaystyle \text{iii) Given progression: }a,\frac{3a^2}{4},\frac{9a^3}{16},\ldots,\quad a\neq0$
$\displaystyle \frac{\frac{3a^2}{4}}{a}=\frac{3a}{4}$
$\displaystyle \frac{\frac{9a^3}{16}}{\frac{3a^2}{4}}=\frac{9a^3}{16}\times\frac{4}{3a^2}=\frac{3a}{4}$
$\displaystyle \therefore \frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{3a}{4}$
$\displaystyle \text{Since the ratio of each term to its preceding term is constant, the progression is a G.P.}$
$\displaystyle \therefore \text{The first term is }a\text{ and the common ratio is }\frac{3a}{4}.$

$\displaystyle \text{iv) Given progression: }\frac{1}{2},\frac{1}{3},\frac{2}{9},\frac{4}{27},\ldots$
$\displaystyle \frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3},\qquad \frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{3},\qquad \frac{\frac{4}{27}}{\frac{2}{9}}=\frac{2}{3}$
$\displaystyle \therefore \frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\frac{2}{3}$
$\displaystyle \text{Since the ratio of each term to its preceding term is constant, the progression is a G.P.}$
$\displaystyle \therefore \text{The first term is }\frac{1}{2}\text{ and the common ratio is }\frac{2}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Show that the sequence defined by }a_n=\frac{2}{3^n},\ n\in N,\text{ is a G.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }a_n=\frac{2}{3^n},\ n\in N.$
$\displaystyle a_1=\frac{2}{3},\qquad a_2=\frac{2}{3^2},\qquad a_3=\frac{2}{3^3},\qquad a_4=\frac{2}{3^4}$
$\displaystyle \frac{a_2}{a_1}=\frac{2/3^2}{2/3}=\frac{1}{3},\qquad \frac{a_3}{a_2}=\frac{2/3^3}{2/3^2}=\frac{1}{3},\qquad \frac{a_4}{a_3}=\frac{2/3^4}{2/3^3}=\frac{1}{3}$
$\displaystyle \therefore \frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}=\frac{1}{3}$
$\displaystyle \text{Since the ratio of each term to its preceding term is constant, the sequence is a G.P.}$
$\displaystyle \therefore \text{The first term is }\frac{2}{3}\text{ and the common ratio is }\frac{1}{3}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find:}$
$\displaystyle \text{i) the }9^{\text{th}}\text{ term of the G.P. }1,4,16,64,\ldots$
$\displaystyle \text{ii) the }10^{\text{th}}\text{ term of the G.P. }-\frac{3}{4},\frac{1}{2},-\frac{1}{3},\frac{2}{9},\ldots$
$\displaystyle \text{iii) the }8^{\text{th}}\text{ term of the G.P. }0.3,0.06,0.012,\ldots$
$\displaystyle \text{iv) the }12^{\text{th}}\text{ term of the G.P. }\frac{1}{a^3x^3},ax,a^5x^5,\ldots$
$\displaystyle \text{v) the }n^{\text{th}}\text{ term of the G.P. }\sqrt{3},\frac{1}{\sqrt{3}},\frac{1}{3\sqrt{3}},\ldots$
$\displaystyle \text{vi) the }10^{\text{th}}\text{ term of the G.P. }\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\ldots$
$\displaystyle \text{Answer:}$

$\displaystyle \text{i) First term }=1,\qquad r=\frac{4}{1}=4$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_9=(1)(4)^{9-1}=4^8=65536$

$\displaystyle \text{ii) First term }=-\frac{3}{4},\qquad r=\frac{\frac12}{-\frac34}=-\frac23$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_{10}=-\frac34\left(-\frac23\right)^9=\frac12\left(\frac23\right)^8$

$\displaystyle \text{iii) First term }=0.3,\qquad r=\frac{0.06}{0.3}=0.2$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_8=(0.3)(0.2)^{8-1}=0.3(0.2)^7$

$\displaystyle \text{iv) First term }=\frac{1}{a^3x^3},\qquad r=\frac{ax}{\frac{1}{a^3x^3}}=a^4x^4$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_{12}=\frac{1}{a^3x^3}(a^4x^4)^{11}=a^{41}x^{41}$

$\displaystyle \text{v) First term }=\sqrt3,\qquad r=\frac{\frac1{\sqrt3}}{\sqrt3}=\frac13$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_n=\sqrt3\left(\frac13\right)^{n-1}$

$\displaystyle \text{vi) First term }=\sqrt2,\qquad r=\frac{\frac1{\sqrt2}}{\sqrt2}=\frac12$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_{10}=\sqrt2\left(\frac12\right)^9=\frac{\sqrt2}{2^9}=\frac{1}{(\sqrt2)^{17}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the }4^{\text{th}}\text{ term from the end of the G.P. }\frac{2}{27},\frac{2}{9},\frac{2}{3},\ldots,162.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{First term }=\frac{2}{27},\qquad r=\frac{\frac{2}{9}}{\frac{2}{27}}=3$
$\displaystyle \text{Last term }l=162$
$\displaystyle \text{On reversing the G.P., the first term becomes }162\text{ and }r=\frac{1}{3}.$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_4=162\left(\frac{1}{3}\right)^{4-1}=\frac{162}{3^3}=6$
$\displaystyle \therefore \text{The }4^{\text{th}}\text{ term from the end is }6.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Which term of the progression }0.004,0.02,0.1,\ldots\text{ is }12.5?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given G.P. }0.004,0.02,0.1,\ldots$
$\displaystyle \text{First term }a=0.004,\qquad r=\frac{0.02}{0.004}=5$
$\displaystyle \text{Let }12.5\text{ be the }n^{\text{th}}\text{ term.}$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore 12.5=(0.004)(5)^{n-1}$
$\displaystyle \Rightarrow 5^{n-1}=\frac{12.5}{0.004}=3125=5^5$
$\displaystyle \therefore n-1=5\Rightarrow n=6$
$\displaystyle \therefore 12.5\text{ is the }6^{\text{th}}\text{ term of the given G.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Which term of the following G.P. is the given number?}$
$\displaystyle \text{i) }\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ldots\text{ is }\frac{1}{512\sqrt{2}}?$
$\displaystyle \text{ii) }2,2\sqrt{2},4,\ldots\text{ is }128?$
$\displaystyle \text{iii) }\sqrt{3},3,3\sqrt{3},\ldots\text{ is }729?$
$\displaystyle \text{iv) }\frac{1}{3},\frac{1}{9},\frac{1}{27},\ldots\text{ is }\frac{1}{19683}?$
$\displaystyle \text{Answer:}$

$\displaystyle \text{i) Given G.P. }\sqrt{2},\frac{1}{\sqrt{2}},\frac{1}{2\sqrt{2}},\frac{1}{4\sqrt{2}},\ldots$
$\displaystyle a=\sqrt{2},\qquad r=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$
$\displaystyle \text{Let }\frac{1}{512\sqrt{2}}\text{ be the }n^{\text{th}}\text{ term.}$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore \frac{1}{512\sqrt{2}}=\sqrt{2}\left(\frac{1}{2}\right)^{n-1}$
$\displaystyle \Rightarrow \frac{1}{1024}=\left(\frac{1}{2}\right)^{n-1}$
$\displaystyle \Rightarrow \left(\frac{1}{2}\right)^{10}=\left(\frac{1}{2}\right)^{n-1}$
$\displaystyle \Rightarrow n-1=10\Rightarrow n=11$
$\displaystyle \therefore \frac{1}{512\sqrt{2}}\text{ is the }11^{\text{th}}\text{ term of the given G.P.}$

$\displaystyle \text{ii) Given G.P. }2,2\sqrt{2},4,\ldots$
$\displaystyle a=2,\qquad r=\frac{2\sqrt{2}}{2}=\sqrt{2}$
$\displaystyle \text{Let }128\text{ be the }n^{\text{th}}\text{ term.}$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore 128=2(\sqrt{2})^{n-1}$
$\displaystyle \Rightarrow 64=(\sqrt{2})^{n-1}$
$\displaystyle \Rightarrow (\sqrt{2})^{12}=(\sqrt{2})^{n-1}$
$\displaystyle \Rightarrow n-1=12\Rightarrow n=13$
$\displaystyle \therefore 128\text{ is the }13^{\text{th}}\text{ term of the given G.P.}$

$\displaystyle \text{iii) Given G.P. }\sqrt{3},3,3\sqrt{3},\ldots$
$\displaystyle a=\sqrt{3},\qquad r=\frac{3}{\sqrt{3}}=\sqrt{3}$
$\displaystyle \text{Let }729\text{ be the }n^{\text{th}}\text{ term.}$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore 729=\sqrt{3}(\sqrt{3})^{n-1}=(\sqrt{3})^n$
$\displaystyle \Rightarrow (\sqrt{3})^{12}=(\sqrt{3})^n$
$\displaystyle \Rightarrow n=12$
$\displaystyle \therefore 729\text{ is the }12^{\text{th}}\text{ term of the given G.P.}$

$\displaystyle \text{iv) Given G.P. }\frac{1}{3},\frac{1}{9},\frac{1}{27},\ldots$
$\displaystyle a=\frac{1}{3},\qquad r=\frac{\frac{1}{9}}{\frac{1}{3}}=\frac{1}{3}$
$\displaystyle \text{Let }\frac{1}{19683}\text{ be the }n^{\text{th}}\text{ term.}$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore \frac{1}{19683}=\frac{1}{3}\left(\frac{1}{3}\right)^{n-1}=\left(\frac{1}{3}\right)^n$
$\displaystyle \Rightarrow \left(\frac{1}{3}\right)^9=\left(\frac{1}{3}\right)^n$
$\displaystyle \Rightarrow n=9$
$\displaystyle \therefore \frac{1}{19683}\text{ is the }9^{\text{th}}\text{ term of the given G.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Which term of the progression }18,-12,8,\ldots\text{ is }\frac{512}{729}?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given G.P. }18,-12,8,\ldots$
$\displaystyle a=18,\qquad r=\frac{-12}{18}=-\frac{2}{3}$
$\displaystyle \text{Let }\frac{512}{729}\text{ be the }n^{\text{th}}\text{ term.}$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore \frac{512}{729}=18\left(-\frac{2}{3}\right)^{n-1}$
$\displaystyle \Rightarrow \left(-\frac{2}{3}\right)^{n-1}=\frac{512}{729\times18}=\frac{256}{6561}$
$\displaystyle \Rightarrow \left(-\frac{2}{3}\right)^{n-1}=\left(-\frac{2}{3}\right)^8$
$\displaystyle \therefore n-1=8\Rightarrow n=9$
$\displaystyle \therefore \frac{512}{729}\text{ is the }9^{\text{th}}\text{ term of the given G.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find the }4^{\text{th}}\text{ term from the end of the G.P. }\frac12,\frac16,\frac1{18},\frac1{54},\ldots,\frac1{4374}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{On reversing the G.P., the first term becomes }\frac1{4374}\text{ and }r=3.$
$\displaystyle \text{We know }a_n=ar^{n-1}$
$\displaystyle \therefore a_4=\frac1{4374}(3)^{4-1}=\frac{27}{4374}=\frac1{162}$
$\displaystyle \therefore \text{The }4^{\text{th}}\text{ term from the end is }\frac1{162}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The }4^{\text{th}}\text{ term of a G.P. is }27\text{ and the }7^{\text{th}}\text{ term is }729.$
$\displaystyle \text{Find the G.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }a_4=27$
$\displaystyle \Rightarrow ar^3=27\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle \text{Given }a_7=729$
$\displaystyle \Rightarrow ar^6=729\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i), we get}$
$\displaystyle \frac{ar^6}{ar^3}=\frac{729}{27}$
$\displaystyle \Rightarrow r^3=27=3^3$
$\displaystyle \therefore r=3$
$\displaystyle \text{Substituting }r=3\text{ in (i), we get}$
$\displaystyle a(3)^3=27$
$\displaystyle \Rightarrow 27a=27$
$\displaystyle \therefore a=1$
$\displaystyle \therefore \text{The required G.P. is }1,3,9,27,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The }7^{\text{th}}\text{ term of a G.P. is }8\text{ times its }4^{\text{th}}\text{ term and the}$
$\displaystyle 5^{\text{th}}\text{ term is }48.\text{ Find the G.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }a_7=8a_4$
$\displaystyle \Rightarrow ar^6=8ar^3$
$\displaystyle \Rightarrow r^3=8=2^3$
$\displaystyle \therefore r=2$
$\displaystyle \text{Also, }a_5=48$
$\displaystyle \Rightarrow ar^4=48$
$\displaystyle \Rightarrow a(2)^4=48$
$\displaystyle \Rightarrow 16a=48$
$\displaystyle \therefore a=3$
$\displaystyle \therefore \text{The required G.P. is }3,6,12,24,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{If the G.P.s }5,10,20,\ldots\text{ and }1280,640,320,\ldots\text{ have equal}$
$\displaystyle n^{\text{th}}\text{ terms, find the value of }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the G.P. }5,10,20,\ldots$
$\displaystyle a=5,\qquad r=\frac{10}{5}=2$
$\displaystyle \text{Its }n^{\text{th}}\text{ term is}$
$\displaystyle T_n=ar^{n-1}=5(2)^{n-1}\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle \text{For the G.P. }1280,640,320,\ldots$
$\displaystyle a=1280,\qquad r=\frac{640}{1280}=\frac{1}{2}$
$\displaystyle \text{Its }n^{\text{th}}\text{ term is}$
$\displaystyle T_n=1280\left(\frac{1}{2}\right)^{n-1}\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Since the }n^{\text{th}}\text{ terms are equal, from (i) and (ii),}$
$\displaystyle 5(2)^{n-1}=1280\left(\frac{1}{2}\right)^{n-1}$
$\displaystyle \Rightarrow 5(2)^{2n-2}=1280$
$\displaystyle \Rightarrow 2^{2n-2}=256=2^8$
$\displaystyle \Rightarrow 2n-2=8$
$\displaystyle \Rightarrow 2n=10$
$\displaystyle \therefore n=5$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{The }5^{\text{th}},8^{\text{th}}\text{ and }11^{\text{th}}\text{ terms of a G.P. are }p,q\text{ and }s$
$\displaystyle \text{respectively. Prove that }q^2=ps.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle a_5=ar^4=p\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle a_8=ar^7=q\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle a_{11}=ar^{10}=s\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle ps=(ar^4)(ar^{10})$
$\displaystyle =a^2r^{14}$
$\displaystyle =(ar^7)^2$
$\displaystyle =q^2$
$\displaystyle \therefore q^2=ps.\text{ Hence proved.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{The }4^{\text{th}}\text{ term of a G.P. with a non-zero common ratio is the square of its}$
$\displaystyle 2^{\text{nd}}\text{ term. If the first term is }-3,\text{ find its }7^{\text{th}}\text{ term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }a_4=(a_2)^2\text{ and }a=-3$
$\displaystyle \Rightarrow ar^3=(ar)^2$
$\displaystyle \Rightarrow ar^3=a^2r^2$
$\displaystyle \Rightarrow ar^2(r-a)=0$
$\displaystyle \text{Since }a=-3\neq0\text{ and }r\neq0,\text{ we get }r=a=-3.$
$\displaystyle \therefore a_7=ar^6$
$\displaystyle =(-3)(-3)^6$
$\displaystyle =(-3)^7$
$\displaystyle =-2187$
$\displaystyle \therefore \text{The }7^{\text{th}}\text{ term of the G.P. is }-2187.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{In a G.P., the }3^{\text{rd}}\text{ term is }24\text{ and the }6^{\text{th}}\text{ term is }192.$
$\displaystyle \text{Find the }10^{\text{th}}\text{ term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle a_3=ar^2=24\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle a_6=ar^5=192\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Dividing (ii) by (i), we get}$
$\displaystyle \frac{ar^5}{ar^2}=\frac{192}{24}$
$\displaystyle \Rightarrow r^3=8=2^3$
$\displaystyle \therefore r=2$
$\displaystyle \text{Substituting }r=2\text{ in (i), we get}$
$\displaystyle a(2)^2=24$
$\displaystyle \Rightarrow 4a=24$
$\displaystyle \therefore a=6$
$\displaystyle a_{10}=ar^9$
$\displaystyle =6(2)^9$
$\displaystyle =6\times512$
$\displaystyle =3072$
$\displaystyle \therefore \text{The }10^{\text{th}}\text{ term is }3072.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{If }a,b,c,d\text{ and }p\text{ are real numbers such that}$
$\displaystyle (a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\leq0,$
$\displaystyle \text{show that }a,b,c\text{ and }d\text{ are in G.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given}$
$\displaystyle (a^2+b^2+c^2)p^2-2(ab+bc+cd)p+(b^2+c^2+d^2)\leq0$
$\displaystyle \Rightarrow a^2p^2+b^2p^2+c^2p^2-2abp-2bcp-2cdp+b^2+c^2+d^2\leq0$
$\displaystyle \Rightarrow (a^2p^2-2abp+b^2)+(b^2p^2-2bcp+c^2)$
$\displaystyle \qquad +(c^2p^2-2cdp+d^2)\leq0$
$\displaystyle \Rightarrow (ap-b)^2+(bp-c)^2+(cp-d)^2\leq0$
$\displaystyle \text{Each term on the left-hand side is non-negative. Therefore, their sum can be at most zero}$
$\displaystyle \text{only when each term is zero.}$
$\displaystyle \therefore (ap-b)^2=0,\qquad (bp-c)^2=0,\qquad (cp-d)^2=0$
$\displaystyle \Rightarrow ap=b,\qquad bp=c,\qquad cp=d$
$\displaystyle \Rightarrow p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\displaystyle \therefore \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\displaystyle \therefore a,b,c\text{ and }d\text{ are in G.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If }\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx},\quad x\neq0,\text{ show that}$
$\displaystyle a,b,c\text{ and }d\text{ are in G.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }\frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$
$\displaystyle \text{Taking the first two ratios,}$
$\displaystyle \frac{a+bx}{a-bx}=\frac{b+cx}{b-cx}$
$\displaystyle \Rightarrow (a+bx)(b-cx)=(a-bx)(b+cx)$
$\displaystyle \Rightarrow ab-acx+b^2x-bcx^2=ab+acx-b^2x-bcx^2$
$\displaystyle \Rightarrow 2x(b^2-ac)=0$
$\displaystyle \text{Since }x\neq0,\text{ we get}$
$\displaystyle b^2=ac\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle \text{Taking the last two ratios,}$
$\displaystyle \frac{b+cx}{b-cx}=\frac{c+dx}{c-dx}$
$\displaystyle \Rightarrow (b+cx)(c-dx)=(b-cx)(c+dx)$
$\displaystyle \Rightarrow bc-bdx+c^2x-cdx^2=bc+bdx-c^2x-cdx^2$
$\displaystyle \Rightarrow 2x(c^2-bd)=0$
$\displaystyle \text{Since }x\neq0,\text{ we get}$
$\displaystyle c^2=bd\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{From (i) and (ii), }b^2=ac\text{ and }c^2=bd.$
$\displaystyle \therefore a,b,c\text{ and }d\text{ are consecutive terms of a G.P.}$
$\displaystyle \text{Hence, }a,b,c\text{ and }d\text{ are in G.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If the }p^{\text{th}}\text{ and }q^{\text{th}}\text{ terms of a G.P. are }q\text{ and }p$
$\displaystyle \text{respectively, where }p\neq q,\text{ show that its }(p+q)^{\text{th}}\text{ term is}$
$\displaystyle \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.$
$\displaystyle a_p=ar^{p-1}=q\qquad\ldots\ldots\ldots\text{(i)}$
$\displaystyle a_q=ar^{q-1}=p\qquad\ldots\ldots\ldots\text{(ii)}$
$\displaystyle \text{Dividing (i) by (ii), we get}$
$\displaystyle \frac{ar^{p-1}}{ar^{q-1}}=\frac{q}{p}$
$\displaystyle \Rightarrow r^{p-q}=\frac{q}{p}$
$\displaystyle \therefore r=\left(\frac{q}{p}\right)^{\frac{1}{p-q}}\qquad\ldots\ldots\ldots\text{(iii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle ar^{q-1}=p$
$\displaystyle \Rightarrow a=\frac{p}{r^{q-1}}$
$\displaystyle =p\left(\frac{p}{q}\right)^{\frac{q-1}{p-q}}\qquad\ldots\ldots\ldots\text{(iv)}$
$\displaystyle \text{Now, the }(p+q)^{\text{th}}\text{ term is}$
$\displaystyle a_{p+q}=ar^{p+q-1}$
$\displaystyle =p\left(\frac{p}{q}\right)^{\frac{q-1}{p-q}}\left(\frac{q}{p}\right)^{\frac{p+q-1}{p-q}}$
$\displaystyle =p\left(\frac{q}{p}\right)^{\frac{p+q-1-(q-1)}{p-q}}$
$\displaystyle =p\left(\frac{q}{p}\right)^{\frac{p}{p-q}}$
$\displaystyle =\frac{q^{\frac{p}{p-q}}}{p^{\frac{q}{p-q}}}$
$\displaystyle =\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}$
$\displaystyle \therefore \text{The }(p+q)^{\text{th}}\text{ term is }\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.$
$\displaystyle \\$