\displaystyle \textbf{Question 1: }\text{If }a,b,c\text{ are positive numbers in G.P., prove that }\log a,\log b,\log c
\displaystyle \text{are in A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in G.P.}
\displaystyle \therefore b^2=ac
\displaystyle \text{Taking logarithms on both sides,}
\displaystyle \log b^2=\log(ac)
\displaystyle \Rightarrow 2\log b=\log a+\log c
\displaystyle \therefore \log a,\log b,\log c\text{ are in A.P.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }a,b,c\text{ are in G.P., prove that }\frac{1}{\log_a m},\frac{1}{\log_b m},\frac{1}{\log_c m}
\displaystyle \text{are in A.P., where }a,b,c,m>0\text{ and }a,b,c,m\neq1.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in G.P.}
\displaystyle \therefore b^2=ac
\displaystyle \text{Taking logarithms to the base }m\text{ on both sides,}
\displaystyle \log_m b^2=\log_m(ac)
\displaystyle \Rightarrow 2\log_m b=\log_m a+\log_m c
\displaystyle \text{Using }\log_m a=\frac{1}{\log_a m},\ \log_m b=\frac{1}{\log_b m}\text{ and }\log_m c=\frac{1}{\log_c m},
\displaystyle \frac{2}{\log_b m}=\frac{1}{\log_a m}+\frac{1}{\log_c m}
\displaystyle \therefore \frac{1}{\log_a m},\frac{1}{\log_b m},\frac{1}{\log_c m}\text{ are in A.P.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find }k\text{ such that }k+9,\ k-6\text{ and }4\text{ form three consecutive terms}
\displaystyle \text{of a G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }k+9,\ k-6\text{ and }4\text{ form three consecutive terms of a G.P.}
\displaystyle \therefore (k-6)^2=4(k+9)
\displaystyle \Rightarrow k^2-12k+36=4k+36
\displaystyle \Rightarrow k^2-16k=0
\displaystyle \Rightarrow k(k-16)=0
\displaystyle \Rightarrow k=0\text{ or }k=16
\displaystyle \text{For }k=0,\text{ the terms are }9,-6,4\text{ and their common ratio is }-\frac{2}{3}.
\displaystyle \text{For }k=16,\text{ the terms are }25,10,4\text{ and their common ratio is }\frac{2}{5}.
\displaystyle \therefore k=0\text{ or }k=16.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Three numbers are in A.P. and their sum is }15.\text{ If }1,3,9\text{ are added to them}
\displaystyle \text{respectively, they form a G.P. Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three numbers in A.P. be }a,\ a+d,\ a+2d.
\displaystyle \text{Given, }a+(a+d)+(a+2d)=15
\displaystyle \Rightarrow 3a+3d=15
\displaystyle \Rightarrow a+d=5\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a+1,\ a+d+3,\ a+2d+9\text{ are in G.P.}
\displaystyle \therefore (a+d+3)^2=(a+1)(a+2d+9)
\displaystyle \text{From (i), }a=5-d.
\displaystyle \therefore (5+3)^2=(5-d+1)(5-d+2d+9)
\displaystyle \Rightarrow 64=(6-d)(14+d)
\displaystyle \Rightarrow 64=84-8d-d^2
\displaystyle \Rightarrow d^2+8d-20=0
\displaystyle \Rightarrow (d-2)(d+10)=0
\displaystyle \Rightarrow d=2\text{ or }d=-10
\displaystyle \text{When }d=2,\ a=5-2=3.
\displaystyle \therefore \text{The three numbers are }3,5,7.
\displaystyle \text{On adding }1,3,9\text{ respectively, we get }4,8,16,\text{ which are in G.P.}
\displaystyle \text{When }d=-10,\ a=5-(-10)=15.
\displaystyle \therefore \text{The three numbers are }15,5,-5.
\displaystyle \text{On adding }1,3,9\text{ respectively, we get }16,8,4,\text{ which are in G.P.}
\displaystyle \therefore \text{The required numbers are }3,5,7\text{ or }15,5,-5.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The sum of three numbers which are consecutive terms of an A.P. is }21.
\displaystyle \text{If the second number is reduced by }1\text{ and the third is increased by }1,\text{ we obtain}
\displaystyle \text{three consecutive terms of a G.P. Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive terms of the A.P. be }a,\ a+d,\ a+2d.
\displaystyle \text{Given, }a+(a+d)+(a+2d)=21
\displaystyle \Rightarrow 3a+3d=21
\displaystyle \Rightarrow a+d=7\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a,\ a+d-1,\ a+2d+1\text{ are in G.P.}
\displaystyle \therefore (a+d-1)^2=a(a+2d+1)
\displaystyle \text{From (i), }d=7-a.
\displaystyle \therefore [a+(7-a)-1]^2=a[a+2(7-a)+1]
\displaystyle \Rightarrow 6^2=a(15-a)
\displaystyle \Rightarrow 36=15a-a^2
\displaystyle \Rightarrow a^2-15a+36=0
\displaystyle \Rightarrow (a-3)(a-12)=0
\displaystyle \Rightarrow a=3\text{ or }a=12
\displaystyle \text{When }a=3,\ d=7-3=4.
\displaystyle \therefore \text{The numbers are }3,7,11.
\displaystyle \text{After the required changes, the numbers become }3,6,12,\text{ which are in G.P.}
\displaystyle \text{When }a=12,\ d=7-12=-5.
\displaystyle \therefore \text{The numbers are }12,7,2.
\displaystyle \text{After the required changes, the numbers become }12,6,3,\text{ which are in G.P.}
\displaystyle \therefore \text{The required numbers are }3,7,11\text{ or }12,7,2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The sum of three numbers }a,b,c\text{ in A.P. is }18.\text{ If }a\text{ and }b\text{ are each}
\displaystyle \text{increased by }4\text{ and }c\text{ is increased by }36,\text{ the new numbers form a G.P.}
\displaystyle \text{Find }a,b,c.
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the A.P. be }a,\ a+d,\ a+2d.
\displaystyle \text{Given, }a+(a+d)+(a+2d)=18
\displaystyle \Rightarrow 3a+3d=18
\displaystyle \Rightarrow a+d=6\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a+4,\ a+d+4,\ a+2d+36\text{ are in G.P.}
\displaystyle \therefore (a+d+4)^2=(a+4)(a+2d+36)
\displaystyle \text{From (i), }a=6-d.
\displaystyle \therefore (6-d+d+4)^2=(6-d+4)[6-d+2d+36]
\displaystyle \Rightarrow 100=(10-d)(42+d)
\displaystyle \Rightarrow 100=420-32d-d^2
\displaystyle \Rightarrow d^2+32d-320=0
\displaystyle \Rightarrow (d+40)(d-8)=0
\displaystyle \Rightarrow d=-40\text{ or }d=8
\displaystyle \text{When }d=-40,\ a=6-(-40)=46.
\displaystyle \therefore a=46,\ b=6,\ c=-34.
\displaystyle \text{After the required increases, the numbers become }50,10,2,\text{ which are in G.P.}
\displaystyle \text{When }d=8,\ a=6-8=-2.
\displaystyle \therefore a=-2,\ b=6,\ c=14.
\displaystyle \text{After the required increases, the numbers become }2,10,50,\text{ which are in G.P.}
\displaystyle \therefore (a,b,c)=(46,6,-34)\text{ or }(-2,6,14).
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The sum of three numbers in G.P. is }56.\text{ If }1,7,21\text{ are subtracted from}
\displaystyle \text{these numbers respectively, we obtain an A.P. Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive terms of the G.P. be }a,\ ar,\ ar^2.
\displaystyle \text{Given, }a+ar+ar^2=56
\displaystyle \Rightarrow a(1+r+r^2)=56\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a-1,\ ar-7,\ ar^2-21\text{ are in A.P.}
\displaystyle \therefore 2(ar-7)=(a-1)+(ar^2-21)
\displaystyle \Rightarrow 2ar-14=a+ar^2-22
\displaystyle \Rightarrow ar^2-2ar+a=8
\displaystyle \Rightarrow a(r-1)^2=8
\displaystyle \Rightarrow a=\frac{8}{(1-r)^2}\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle \frac{8}{(1-r)^2}(1+r+r^2)=56
\displaystyle \Rightarrow 1+r+r^2=7(1-r)^2
\displaystyle \Rightarrow 1+r+r^2=7-14r+7r^2
\displaystyle \Rightarrow 6r^2-15r+6=0
\displaystyle \Rightarrow 2r^2-5r+2=0
\displaystyle \Rightarrow (r-2)(2r-1)=0
\displaystyle \Rightarrow r=2\text{ or }r=\frac{1}{2}
\displaystyle \text{When }r=2,\ a=\frac{8}{(1-2)^2}=8.
\displaystyle \therefore \text{The numbers are }8,16,32.
\displaystyle \text{After subtracting }1,7,21,\text{ respectively, we get }7,9,11,\text{ which are in A.P.}
\displaystyle \text{When }r=\frac{1}{2},\ a=\frac{8}{\left(1-\frac{1}{2}\right)^2}=32.
\displaystyle \therefore \text{The numbers are }32,16,8.
\displaystyle \text{After subtracting }1,7,21,\text{ respectively, we get }31,9,-13,\text{ which are in A.P.}
\displaystyle \therefore \text{The required numbers are }8,16,32\text{ or }32,16,8.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }a,b,c\text{ are in G.P., prove that:}
\displaystyle \text{i) }a(b^2+c^2)=c(a^2+b^2)
\displaystyle \text{ii) }a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=a^3+b^3+c^3
\displaystyle \text{iii) }\frac{(a+b+c)^2}{a^2+b^2+c^2}=\frac{a+b+c}{a-b+c}
\displaystyle \text{iv) }\frac{1}{a^2-b^2}+\frac{1}{b^2}=\frac{1}{b^2-c^2}
\displaystyle \text{v) }(a+2b+2c)(a-2b+2c)=a^2+4c^2
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,b,c\text{ are in G.P.,}
\displaystyle b^2=ac
\displaystyle \text{i) LHS}=a(b^2+c^2)
\displaystyle =ab^2+ac^2
\displaystyle =a(ac)+b^2c
\displaystyle =a^2c+b^2c
\displaystyle =c(a^2+b^2)=\text{RHS. Hence proved.}
\displaystyle \text{ii) Assuming }a,b,c\neq0,
\displaystyle \text{LHS}=a^2b^2c^2\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)
\displaystyle =\frac{b^2c^2}{a}+\frac{a^2c^2}{b}+\frac{a^2b^2}{c}
\displaystyle =\frac{(ac)c^2}{a}+\frac{(b^2)^2}{b}+\frac{a^2(ac)}{c}
\displaystyle =c^3+b^3+a^3
\displaystyle =a^3+b^3+c^3=\text{RHS. Hence proved.}
\displaystyle \text{iii) LHS}=\frac{(a+b+c)^2}{a^2+b^2+c^2}
\displaystyle =\frac{(a+b+c)^2}{a^2-b^2+c^2+2b^2}
\displaystyle =\frac{(a+b+c)^2}{a^2-b^2+c^2+2ac}
\displaystyle =\frac{(a+b+c)^2}{(a+c)^2-b^2}
\displaystyle =\frac{(a+b+c)^2}{(a+b+c)(a-b+c)}
\displaystyle =\frac{a+b+c}{a-b+c}=\text{RHS. Hence proved.}
\displaystyle \text{iv) Assuming }a^2\neq b^2,\ b\neq0\text{ and }b^2\neq c^2,
\displaystyle \text{LHS}=\frac{1}{a^2-b^2}+\frac{1}{b^2}
\displaystyle =\frac{b^2+a^2-b^2}{b^2(a^2-b^2)}
\displaystyle =\frac{a^2}{a^2b^2-b^4}
\displaystyle =\frac{a^2}{a^2(ac)-(ac)^2}
\displaystyle =\frac{a^2}{a^2c(a-c)}
\displaystyle =\frac{1}{c(a-c)}
\displaystyle =\frac{1}{ac-c^2}
\displaystyle =\frac{1}{b^2-c^2}=\text{RHS. Hence proved.}
\displaystyle \text{v) LHS}=(a+2b+2c)(a-2b+2c)
\displaystyle =[(a+2c)+2b][(a+2c)-2b]
\displaystyle =(a+2c)^2-4b^2
\displaystyle =a^2+4ac+4c^2-4b^2
\displaystyle =a^2+4ac+4c^2-4ac
\displaystyle =a^2+4c^2=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }a,b,c,d\text{ are four consecutive terms of a G.P., prove that:}
\displaystyle \text{i) }\frac{ab-cd}{b^2-c^2}=\frac{a+c}{b}
\displaystyle \text{ii) }(a+b+c+d)^2=(a+b)^2+2(b+c)^2+(c+d)^2
\displaystyle \text{iii) }(b+c)(b+d)=(c+a)(c+d)
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,b,c,d\text{ are four consecutive terms of a G.P., let their common ratio be }r.
\displaystyle \therefore b=ar,\qquad c=ar^2,\qquad d=ar^3
\displaystyle \text{Hence, }b^2=ac,\qquad c^2=bd,\qquad ad=bc.
\displaystyle \text{i) LHS}=\frac{ab-cd}{b^2-c^2}
\displaystyle =\frac{a(ar)-(ar^2)(ar^3)}{(ar)^2-(ar^2)^2}
\displaystyle =\frac{a^2r(1-r^4)}{a^2r^2(1-r^2)}
\displaystyle =\frac{(1-r^2)(1+r^2)}{r(1-r^2)}
\displaystyle =\frac{1+r^2}{r}
\displaystyle =\frac{a+ar^2}{ar}
\displaystyle =\frac{a+c}{b}=\text{RHS. Hence proved.}
\displaystyle \text{ii) LHS}=(a+b+c+d)^2
\displaystyle =[(a+b)+(c+d)]^2
\displaystyle =(a+b)^2+2(a+b)(c+d)+(c+d)^2
\displaystyle =(a+b)^2+2(ac+ad+bc+bd)+(c+d)^2
\displaystyle =(a+b)^2+2(b^2+bc+bc+c^2)+(c+d)^2
\displaystyle =(a+b)^2+2(b^2+2bc+c^2)+(c+d)^2
\displaystyle =(a+b)^2+2(b+c)^2+(c+d)^2=\text{RHS. Hence proved.}
\displaystyle \text{iii) LHS}=(b+c)(b+d)
\displaystyle =b^2+bc+bd+cd
\displaystyle =ac+ad+c^2+cd
\displaystyle =c(a+c)+d(a+c)
\displaystyle =(a+c)(c+d)
\displaystyle =(c+a)(c+d)=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }a,b,c\text{ are in G.P., prove that the following are also in G.P.:}
\displaystyle \text{i) }a^2,b^2,c^2\qquad\text{ii) }a^3,b^3,c^3
\displaystyle \text{iii) }a^2+b^2,\ ab+bc,\ b^2+c^2
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in G.P.}
\displaystyle \therefore b^2=ac
\displaystyle \text{i) Squaring both sides of }b^2=ac,
\displaystyle (b^2)^2=(ac)^2
\displaystyle \Rightarrow (b^2)^2=a^2c^2
\displaystyle \therefore a^2,b^2,c^2\text{ are in G.P.}
\displaystyle \text{ii) Cubing both sides of }b^2=ac,
\displaystyle (b^2)^3=(ac)^3
\displaystyle \Rightarrow b^6=a^3c^3
\displaystyle \Rightarrow (b^3)^2=a^3c^3
\displaystyle \therefore a^3,b^3,c^3\text{ are in G.P.}
\displaystyle \text{iii) We have,}
\displaystyle (ab+bc)^2=b^2(a+c)^2
\displaystyle =ac(a+c)^2
\displaystyle =ac(a^2+2ac+c^2)
\displaystyle =a^3c+2a^2c^2+ac^3
\displaystyle \text{Also,}
\displaystyle (a^2+b^2)(b^2+c^2)
\displaystyle =(a^2+ac)(ac+c^2)
\displaystyle =a(a+c)\cdot c(a+c)
\displaystyle =ac(a+c)^2
\displaystyle \therefore (ab+bc)^2=(a^2+b^2)(b^2+c^2)
\displaystyle \therefore a^2+b^2,\ ab+bc,\ b^2+c^2\text{ are in G.P.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }a,b,c,d\text{ are four consecutive terms of a G.P., prove that:}
\displaystyle \text{i) }a^2+b^2,\ b^2+c^2,\ c^2+d^2\text{ are in G.P.}
\displaystyle \text{ii) }a^2-b^2,\ b^2-c^2,\ c^2-d^2\text{ are in G.P.}
\displaystyle \text{iii) }\frac{1}{a^2+b^2},\ \frac{1}{b^2+c^2},\ \frac{1}{c^2+d^2}\text{ are in G.P.}
\displaystyle \text{iv) }a^2+b^2+c^2,\ ab+bc+cd,\ b^2+c^2+d^2\text{ are in G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the four consecutive terms of the G.P. be }a,\ ar,\ ar^2,\ ar^3.
\displaystyle \therefore b=ar,\qquad c=ar^2,\qquad d=ar^3
\displaystyle \text{i) }a^2+b^2=a^2+a^2r^2=a^2(1+r^2)
\displaystyle b^2+c^2=a^2r^2+a^2r^4=a^2r^2(1+r^2)
\displaystyle c^2+d^2=a^2r^4+a^2r^6=a^2r^4(1+r^2)
\displaystyle \therefore \frac{b^2+c^2}{a^2+b^2}=r^2
\displaystyle \text{and }\frac{c^2+d^2}{b^2+c^2}=r^2.
\displaystyle \therefore a^2+b^2,\ b^2+c^2,\ c^2+d^2\text{ are in G.P.}
\displaystyle \text{ii) }a^2-b^2=a^2-a^2r^2=a^2(1-r^2)
\displaystyle b^2-c^2=a^2r^2-a^2r^4=a^2r^2(1-r^2)
\displaystyle c^2-d^2=a^2r^4-a^2r^6=a^2r^4(1-r^2)
\displaystyle \therefore \frac{b^2-c^2}{a^2-b^2}=r^2
\displaystyle \text{and }\frac{c^2-d^2}{b^2-c^2}=r^2.
\displaystyle \therefore a^2-b^2,\ b^2-c^2,\ c^2-d^2\text{ are in G.P.}
\displaystyle \text{iii) Assuming the denominators are non-zero,}
\displaystyle \frac{1}{a^2+b^2}=\frac{1}{a^2(1+r^2)}
\displaystyle \frac{1}{b^2+c^2}=\frac{1}{a^2r^2(1+r^2)}
\displaystyle \frac{1}{c^2+d^2}=\frac{1}{a^2r^4(1+r^2)}
\displaystyle \therefore \frac{\frac{1}{b^2+c^2}}{\frac{1}{a^2+b^2}}=\frac{1}{r^2}
\displaystyle \text{and }\frac{\frac{1}{c^2+d^2}}{\frac{1}{b^2+c^2}}=\frac{1}{r^2}.
\displaystyle \therefore \frac{1}{a^2+b^2},\ \frac{1}{b^2+c^2},\ \frac{1}{c^2+d^2}\text{ are in G.P.}
\displaystyle \text{iv) }a^2+b^2+c^2=a^2+a^2r^2+a^2r^4
\displaystyle =a^2(1+r^2+r^4)
\displaystyle ab+bc+cd=a^2r+a^2r^3+a^2r^5
\displaystyle =a^2r(1+r^2+r^4)
\displaystyle b^2+c^2+d^2=a^2r^2+a^2r^4+a^2r^6
\displaystyle =a^2r^2(1+r^2+r^4)
\displaystyle \therefore \frac{ab+bc+cd}{a^2+b^2+c^2}=r
\displaystyle \text{and }\frac{b^2+c^2+d^2}{ab+bc+cd}=r.
\displaystyle \therefore a^2+b^2+c^2,\ ab+bc+cd,\ b^2+c^2+d^2\text{ are in G.P.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }a-b,\ b-c,\ c-a\text{ are in G.P., prove that}
\displaystyle (a+b+c)^2=3(ab+bc+ca).
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a-b,\ b-c,\ c-a\text{ are in G.P.}
\displaystyle \therefore (b-c)^2=(a-b)(c-a)
\displaystyle \Rightarrow b^2-2bc+c^2=ac-a^2-bc+ab
\displaystyle \Rightarrow a^2+b^2+c^2=ab+bc+ca\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{LHS}=(a+b+c)^2
\displaystyle =a^2+b^2+c^2+2ab+2bc+2ca
\displaystyle =ab+bc+ca+2ab+2bc+2ca\qquad\text{[Using (i)]}
\displaystyle =3(ab+bc+ca)
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }a,b,c\text{ are in G.P., prove that}
\displaystyle \frac{a^2+ab+b^2}{bc+ca+ab}=\frac{b+a}{c+b}.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in G.P.}
\displaystyle \therefore b^2=ac
\displaystyle \text{Assuming the denominators are non-zero,}
\displaystyle \text{LHS}=\frac{a^2+ab+b^2}{bc+ca+ab}
\displaystyle =\frac{a^2+ab+ac}{bc+b^2+ab}
\displaystyle =\frac{a(a+b+c)}{b(a+b+c)}
\displaystyle =\frac{a}{b}
\displaystyle \text{Also, RHS}=\frac{a+b}{b+c}
\displaystyle =\frac{a(a+b)}{a(b+c)}
\displaystyle =\frac{a(a+b)}{ab+ac}
\displaystyle =\frac{a(a+b)}{ab+b^2}
\displaystyle =\frac{a(a+b)}{b(a+b)}
\displaystyle =\frac{a}{b}
\displaystyle \therefore \text{LHS}=\text{RHS. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If the }4^{\mathrm{th}},10^{\mathrm{th}}\text{ and }16^{\mathrm{th}}\text{ terms of a G.P. are }x,y\text{ and }z
\displaystyle \text{respectively, prove that }x,y,z\text{ are in G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term and common ratio of the G.P. be }a\text{ and }r\text{ respectively.}
\displaystyle \text{Given, }x=ar^3,\qquad y=ar^9,\qquad z=ar^{15}
\displaystyle \therefore y^2=(ar^9)^2
\displaystyle =a^2r^{18}
\displaystyle \text{Also, }xz=(ar^3)(ar^{15})
\displaystyle =a^2r^{18}
\displaystyle \therefore y^2=xz
\displaystyle \therefore x,y,z\text{ are in G.P.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }a,b,c\text{ are in A.P. and }a,b,d\text{ are in G.P., prove that}
\displaystyle a,\ a-b,\ d-c\text{ are in G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in A.P.}
\displaystyle \therefore 2b=a+c\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a,b,d\text{ are in G.P.}
\displaystyle \therefore b^2=ad\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{To prove that }a,\ a-b,\ d-c\text{ are in G.P., it is sufficient to prove that}
\displaystyle (a-b)^2=a(d-c).
\displaystyle \text{LHS}=(a-b)^2
\displaystyle =a^2-2ab+b^2
\displaystyle =a^2-a(a+c)+b^2\qquad\text{[Using (i)]}
\displaystyle =a^2-a^2-ac+ad\qquad\text{[Using (ii)]}
\displaystyle =ad-ac
\displaystyle =a(d-c)
\displaystyle =\text{RHS.}
\displaystyle \therefore a,\ a-b,\ d-c\text{ are in G.P. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If the }p^{\mathrm{th}},q^{\mathrm{th}},r^{\mathrm{th}}\text{ and }s^{\mathrm{th}}\text{ terms of a non-constant A.P.}
\displaystyle \text{are in G.P., prove that }p-q,\ q-r,\ r-s\text{ are in G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term and common difference of the A.P. be }a\text{ and }d,\text{ where }d\neq0.
\displaystyle \text{Since its }p^{\mathrm{th}},q^{\mathrm{th}},r^{\mathrm{th}}\text{ and }s^{\mathrm{th}}\text{ terms are in G.P., let them be}
\displaystyle A,\ AR,\ AR^2,\ AR^3\text{ respectively.}
\displaystyle \therefore a+(p-1)d=A\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle a+(q-1)d=AR\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle a+(r-1)d=AR^2\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle a+(s-1)d=AR^3\qquad\ldots\ldots\ldots\text{(iv)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle [a+(p-1)d]-[a+(q-1)d]=A-AR
\displaystyle \Rightarrow (p-q)d=A(1-R)\qquad\ldots\ldots\ldots\text{(v)}
\displaystyle \text{Subtracting (iii) from (ii),}
\displaystyle [a+(q-1)d]-[a+(r-1)d]=AR-AR^2
\displaystyle \Rightarrow (q-r)d=AR(1-R)\qquad\ldots\ldots\ldots\text{(vi)}
\displaystyle \text{Subtracting (iv) from (iii),}
\displaystyle [a+(r-1)d]-[a+(s-1)d]=AR^2-AR^3
\displaystyle \Rightarrow (r-s)d=AR^2(1-R)\qquad\ldots\ldots\ldots\text{(vii)}
\displaystyle \text{Squaring (vi),}
\displaystyle (q-r)^2d^2=A^2R^2(1-R)^2
\displaystyle =[A(1-R)][AR^2(1-R)]
\displaystyle =[(p-q)d][(r-s)d]\qquad\text{[Using (v) and (vii)]}
\displaystyle \Rightarrow (q-r)^2d^2=(p-q)(r-s)d^2
\displaystyle \Rightarrow (q-r)^2=(p-q)(r-s)\qquad[d\neq0]
\displaystyle \therefore p-q,\ q-r,\ r-s\text{ are in G.P.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }\frac{1}{a+b},\frac{1}{2b},\frac{1}{b+c}\text{ are three consecutive terms of an A.P.,}
\displaystyle \text{prove that }a,b,c\text{ are three consecutive terms of a G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\frac{1}{a+b},\frac{1}{2b},\frac{1}{b+c}\text{ are three consecutive terms of an A.P.}
\displaystyle \text{Assuming }a+b\neq0,\ b\neq0\text{ and }b+c\neq0,
\displaystyle 2\left(\frac{1}{2b}\right)=\frac{1}{a+b}+\frac{1}{b+c}
\displaystyle \Rightarrow \frac{1}{b}=\frac{(b+c)+(a+b)}{(a+b)(b+c)}
\displaystyle \Rightarrow \frac{1}{b}=\frac{a+2b+c}{(a+b)(b+c)}
\displaystyle \Rightarrow (a+b)(b+c)=b(a+2b+c)
\displaystyle \Rightarrow ab+ac+b^2+bc=ab+2b^2+bc
\displaystyle \Rightarrow ac=b^2
\displaystyle \therefore a,b,c\text{ are three consecutive terms of a G.P.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }x^a=x^{\frac{b}{2}}z^{\frac{b}{2}}=z^c\neq1,\text{ where }x,z>0\text{ and }a,b,c\neq0,
\displaystyle \text{prove that }\frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }x^a=x^{\frac{b}{2}}z^{\frac{b}{2}}=z^c.
\displaystyle \text{Taking logarithms,}
\displaystyle a\log x=\frac{b}{2}(\log x+\log z)=c\log z.
\displaystyle \text{Let the common non-zero value be }K.
\displaystyle \therefore a\log x=K\qquad\text{and}\qquad c\log z=K
\displaystyle \Rightarrow \log x=\frac{K}{a}\qquad\text{and}\qquad\log z=\frac{K}{c}
\displaystyle \text{Also, }\frac{b}{2}(\log x+\log z)=K.
\displaystyle \therefore \frac{b}{2}\left(\frac{K}{a}+\frac{K}{c}\right)=K
\displaystyle \Rightarrow \frac{b}{2}\left(\frac{1}{a}+\frac{1}{c}\right)=1\qquad[K\neq0]
\displaystyle \Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b}
\displaystyle \therefore \frac{1}{a},\frac{1}{b},\frac{1}{c}\text{ are in A.P.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }a,b,c\text{ are in A.P., }b,c,d\text{ are in G.P. and }\frac{1}{c},\frac{1}{d},\frac{1}{e}
\displaystyle \text{are in A.P., prove that }a,c,e\text{ are in G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in A.P.}
\displaystyle \therefore 2b=a+c\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }b,c,d\text{ are in G.P.}
\displaystyle \therefore c^2=bd\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Also, }\frac{1}{c},\frac{1}{d},\frac{1}{e}\text{ are in A.P., where }c,d,e\neq0.
\displaystyle \therefore \frac{2}{d}=\frac{1}{c}+\frac{1}{e}
\displaystyle \Rightarrow \frac{2}{d}=\frac{c+e}{ce}
\displaystyle \Rightarrow d(c+e)=2ce\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \text{Multiplying (iii) by }b,
\displaystyle bd(c+e)=2bce
\displaystyle \Rightarrow c^2(c+e)=2bce\qquad\text{[Using (ii)]}
\displaystyle \Rightarrow c(c+e)=2be\qquad[c\neq0]
\displaystyle \Rightarrow c(c+e)=e(a+c)\qquad\text{[Using (i)]}
\displaystyle \Rightarrow c^2+ce=ae+ce
\displaystyle \Rightarrow c^2=ae
\displaystyle \therefore a,c,e\text{ are in G.P.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }a,b,c\text{ are in A.P., and }a,x,b\text{ and }b,y,c\text{ are in G.P., show that}
\displaystyle x^2,b^2,y^2\text{ are in A.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in A.P.}
\displaystyle \therefore 2b=a+c\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a,x,b\text{ are in G.P.}
\displaystyle \therefore x^2=ab\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Similarly, }b,y,c\text{ are in G.P.}
\displaystyle \therefore y^2=bc\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \text{Adding (ii) and (iii),}
\displaystyle x^2+y^2=ab+bc
\displaystyle =b(a+c)
\displaystyle =b(2b)\qquad\text{[Using (i)]}
\displaystyle =2b^2
\displaystyle \therefore 2b^2=x^2+y^2
\displaystyle \therefore x^2,b^2,y^2\text{ are in A.P.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }a,b,c\text{ are in A.P. and }a,b,d\text{ are in G.P., show that}
\displaystyle a,\ a-b,\ d-c\text{ are in G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a,b,c\text{ are in A.P.}
\displaystyle \therefore 2b=a+c\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a,b,d\text{ are in G.P.}
\displaystyle \therefore b^2=ad\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{To prove that }a,\ a-b,\ d-c\text{ are in G.P., it is sufficient to prove that}
\displaystyle (a-b)^2=a(d-c).
\displaystyle \text{LHS}=(a-b)^2
\displaystyle =a^2-2ab+b^2
\displaystyle =a^2-a(a+c)+ad\qquad\text{[Using (i) and (ii)]}
\displaystyle =a^2-a^2-ac+ad
\displaystyle =a(d-c)
\displaystyle =\text{RHS.}
\displaystyle \therefore a,\ a-b,\ d-c\text{ are in G.P. Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }a,b,c\text{ are three distinct real numbers in G.P. and }a+b+c=xb,
\displaystyle \text{prove that either }x<-1\text{ or }x>3.
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ be the common ratio of the G.P.}
\displaystyle \therefore b=ar\qquad\text{and}\qquad c=ar^2
\displaystyle \text{Given, }a+b+c=xb
\displaystyle \Rightarrow a+ar+ar^2=xar
\displaystyle \Rightarrow ar^2+(1-x)ar+a=0
\displaystyle \text{Since }a,b,c\text{ are distinct, }a\neq0.
\displaystyle \therefore r^2+(1-x)r+1=0
\displaystyle \text{Since }r\text{ is real, the discriminant must be non-negative.}
\displaystyle \therefore (1-x)^2-4\geq0
\displaystyle \Rightarrow (1-x-2)(1-x+2)\geq0
\displaystyle \Rightarrow (-x-1)(3-x)\geq0
\displaystyle \Rightarrow (x+1)(x-3)\geq0
\displaystyle \Rightarrow x\leq-1\text{ or }x\geq3
\displaystyle \text{If }x=3,\text{ then }r^2-2r+1=0
\displaystyle \Rightarrow (r-1)^2=0
\displaystyle \Rightarrow r=1,\text{ which gives }a=b=c,\text{ a contradiction.}
\displaystyle \text{If }x=-1,\text{ then }r^2+2r+1=0
\displaystyle \Rightarrow (r+1)^2=0
\displaystyle \Rightarrow r=-1,\text{ which gives }a=c,\text{ a contradiction.}
\displaystyle \therefore x\neq-1\text{ and }x\neq3.
\displaystyle \therefore x<-1\text{ or }x>3.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If the }p^{\mathrm{th}},q^{\mathrm{th}}\text{ and }r^{\mathrm{th}}\text{ terms of an A.P. and a G.P. are}
\displaystyle a,b,c\text{ respectively, prove that }a^{b-c}b^{c-a}c^{a-b}=1,\text{ where }a,b,c>0.
\displaystyle \text{Answer:}
\displaystyle \text{Let }A\text{ and }d\text{ be the first term and common difference of the A.P. respectively.}
\displaystyle \therefore A+(p-1)d=a\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle A+(q-1)d=b\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle A+(r-1)d=c\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle (p-q)d=a-b\qquad\ldots\ldots\ldots\text{(iv)}
\displaystyle \text{Subtracting (iii) from (ii),}
\displaystyle (q-r)d=b-c\qquad\ldots\ldots\ldots\text{(v)}
\displaystyle \text{Subtracting (i) from (iii),}
\displaystyle (r-p)d=c-a\qquad\ldots\ldots\ldots\text{(vi)}
\displaystyle \text{Let }A'\text{ and }R\text{ be the first term and common ratio of the G.P. respectively.}
\displaystyle \therefore A'R^{p-1}=a\qquad\ldots\ldots\ldots\text{(vii)}
\displaystyle A'R^{q-1}=b\qquad\ldots\ldots\ldots\text{(viii)}
\displaystyle A'R^{r-1}=c\qquad\ldots\ldots\ldots\text{(ix)}
\displaystyle \text{LHS}=a^{b-c}b^{c-a}c^{a-b}
\displaystyle =[A'R^{p-1}]^{b-c}[A'R^{q-1}]^{c-a}[A'R^{r-1}]^{a-b}
\displaystyle =[A'R^{p-1}]^{(q-r)d}[A'R^{q-1}]^{(r-p)d}[A'R^{r-1}]^{(p-q)d}
\displaystyle ={A'}^{[(q-r)+(r-p)+(p-q)]d}
\displaystyle \qquad\times R^{[(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)]d}
\displaystyle ={A'}^0R^0
\displaystyle =1
\displaystyle =\text{RHS. Hence proved.}
\displaystyle \\

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