\displaystyle \textbf{Question 1: }\text{Find three numbers in G.P. whose sum is }65\text{ and whose product is }3375.
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=3375
\displaystyle \Rightarrow a^3=3375
\displaystyle \Rightarrow a^3=15^3
\displaystyle \therefore a=15
\displaystyle \text{Also, }\frac{a}{r}+a+ar=65
\displaystyle \Rightarrow \frac{15}{r}+15+15r=65
\displaystyle \text{Multiplying throughout by }r,
\displaystyle 15+15r+15r^2=65r
\displaystyle \Rightarrow 15r^2-50r+15=0
\displaystyle \Rightarrow 3r^2-10r+3=0
\displaystyle \Rightarrow (3r-1)(r-3)=0
\displaystyle \therefore r=\frac{1}{3}\text{ or }r=3
\displaystyle \text{When }r=3,\text{ the terms are }\frac{15}{3},15,15(3),\text{ i.e. }5,15,45.
\displaystyle \text{When }r=\frac{1}{3},\text{ the terms are }45,15,5,\text{ which are the same numbers in reverse order.}
\displaystyle \therefore \text{The required three numbers are }5,15\text{ and }45.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find three numbers in G.P. whose sum is }38\text{ and whose product is }1728.
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=1728
\displaystyle \Rightarrow a^3=1728
\displaystyle \Rightarrow a^3=12^3
\displaystyle \therefore a=12
\displaystyle \text{Also, }\frac{a}{r}+a+ar=38
\displaystyle \Rightarrow \frac{12}{r}+12+12r=38
\displaystyle \text{Multiplying throughout by }r,
\displaystyle 12+12r+12r^2=38r
\displaystyle \Rightarrow 12r^2-26r+12=0
\displaystyle \Rightarrow 6r^2-13r+6=0
\displaystyle \Rightarrow (3r-2)(2r-3)=0
\displaystyle \therefore r=\frac{2}{3}\text{ or }r=\frac{3}{2}
\displaystyle \text{When }r=\frac{3}{2},\text{ the terms are }\frac{12}{3/2},12,12\left(\frac{3}{2}\right),\text{ i.e. }8,12,18.
\displaystyle \text{When }r=\frac{2}{3},\text{ the terms are }18,12,8,\text{ which are the same numbers in reverse order.}
\displaystyle \therefore \text{The required three numbers are }8,12\text{ and }18.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The sum of the first three terms of a G.P. is }\frac{13}{12}\text{ and their product is }-1.
\displaystyle \text{Find the G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=-1
\displaystyle \Rightarrow a^3=-1
\displaystyle \therefore a=-1
\displaystyle \text{Also, }\frac{a}{r}+a+ar=\frac{13}{12}
\displaystyle \Rightarrow -\frac{1}{r}-1-r=\frac{13}{12}
\displaystyle \text{Multiplying throughout by }12r,
\displaystyle -12-12r-12r^2=13r
\displaystyle \Rightarrow 12r^2+25r+12=0
\displaystyle \Rightarrow 12r^2+16r+9r+12=0
\displaystyle \Rightarrow 4r(3r+4)+3(3r+4)=0
\displaystyle \Rightarrow (3r+4)(4r+3)=0
\displaystyle \therefore r=-\frac{4}{3}\text{ or }r=-\frac{3}{4}
\displaystyle \text{When }r=-\frac{4}{3},\text{ the terms are }\frac{-1}{-4/3},-1,(-1)\left(-\frac{4}{3}\right).
\displaystyle \text{Thus, the G.P. is }\frac{3}{4},-1,\frac{4}{3},\ldots
\displaystyle \text{When }r=-\frac{3}{4},\text{ the G.P. is }\frac{4}{3},-1,\frac{3}{4},\ldots,
\displaystyle \text{which consists of the same three terms in reverse order.}
\displaystyle \therefore \text{The required G.P. is }\frac{3}{4},-1,\frac{4}{3},\ldots
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The product of the first three terms of a G.P. is }125\text{ and the sum of their}
\displaystyle \text{products taken in pairs is }87\frac{1}{2}.\text{ Find the terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=125
\displaystyle \Rightarrow a^3=125
\displaystyle \therefore a=5
\displaystyle \text{Also, }\frac{a}{r}\times a+a\times ar+ar\times\frac{a}{r}=87\frac{1}{2}=\frac{175}{2}
\displaystyle \Rightarrow \frac{a^2}{r}+a^2r+a^2=\frac{175}{2}
\displaystyle \Rightarrow \frac{25}{r}+25r+25=\frac{175}{2}
\displaystyle \Rightarrow \frac{1}{r}+r+1=\frac{7}{2}
\displaystyle \text{Multiplying throughout by }2r,
\displaystyle 2+2r^2+2r=7r
\displaystyle \Rightarrow 2r^2-5r+2=0
\displaystyle \Rightarrow (2r-1)(r-2)=0
\displaystyle \therefore r=\frac{1}{2}\text{ or }r=2
\displaystyle \text{When }r=2,\text{ the terms are }\frac{5}{2},5,10.
\displaystyle \text{When }r=\frac{1}{2},\text{ the terms are }10,5,\frac{5}{2},\text{ which are the same in reverse order.}
\displaystyle \therefore \text{The required three terms are }\frac{5}{2},5\text{ and }10.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The sum of the first three terms of a G.P. is }\frac{39}{10}\text{ and their product is }1.
\displaystyle \text{Find the common ratio and the terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=1
\displaystyle \Rightarrow a^3=1
\displaystyle \therefore a=1
\displaystyle \text{Also, }\frac{a}{r}+a+ar=\frac{39}{10}
\displaystyle \Rightarrow \frac{1}{r}+1+r=\frac{39}{10}
\displaystyle \text{Multiplying throughout by }10r,
\displaystyle 10+10r+10r^2=39r
\displaystyle \Rightarrow 10r^2-29r+10=0
\displaystyle \Rightarrow 10r^2-25r-4r+10=0
\displaystyle \Rightarrow 5r(2r-5)-2(2r-5)=0
\displaystyle \Rightarrow (2r-5)(5r-2)=0
\displaystyle \therefore r=\frac{5}{2}\text{ or }r=\frac{2}{5}
\displaystyle \text{When }r=\frac{5}{2},\text{ the terms are }\frac{1}{5/2},1,\frac{5}{2},\text{ i.e. }\frac{2}{5},1,\frac{5}{2}.
\displaystyle \text{When }r=\frac{2}{5},\text{ the terms are }\frac{5}{2},1,\frac{2}{5},
\displaystyle \text{which are the same terms in reverse order.}
\displaystyle \therefore \text{The common ratio is }\frac{5}{2}\text{ or }\frac{2}{5}.
\displaystyle \therefore \text{The required terms are }\frac{2}{5},1\text{ and }\frac{5}{2}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The sum of three numbers in G.P. is }14.\text{ If each of the first two terms is}
\displaystyle \text{increased by }1\text{ and the third term is decreased by }1,\text{ the resulting numbers are in A.P.}
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Since their sum is }14,
\displaystyle \frac{a}{r}+a+ar=14\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{The new terms }\frac{a}{r}+1,\ a+1,\ ar-1\text{ are in A.P.}
\displaystyle \therefore 2(a+1)=\left(\frac{a}{r}+1\right)+(ar-1)
\displaystyle \Rightarrow 2(a+1)=\frac{a}{r}+ar
\displaystyle \text{From (i), }\frac{a}{r}+ar=14-a.
\displaystyle \therefore 2(a+1)=14-a
\displaystyle \Rightarrow 2a+2=14-a
\displaystyle \Rightarrow 3a=12
\displaystyle \therefore a=4
\displaystyle \text{Substituting }a=4\text{ in (i), we get}
\displaystyle \frac{4}{r}+4+4r=14
\displaystyle \Rightarrow \frac{2}{r}+2+2r=7
\displaystyle \text{Multiplying throughout by }r,
\displaystyle 2+2r+2r^2=7r
\displaystyle \Rightarrow 2r^2-5r+2=0
\displaystyle \Rightarrow (r-2)(2r-1)=0
\displaystyle \therefore r=2\text{ or }r=\frac{1}{2}
\displaystyle \text{When }r=2,\text{ the terms are }\frac{4}{2},4,4(2),\text{ i.e. }2,4,8.
\displaystyle \text{When }r=\frac{1}{2},\text{ the terms are }8,4,2,\text{ which are the same in reverse order.}
\displaystyle \therefore \text{The required numbers are }2,4\text{ and }8.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The product of three numbers in G.P. is }216.\text{ If }2,8\text{ and }6\text{ are added}
\displaystyle \text{to them respectively, the resulting numbers are in A.P. Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=216
\displaystyle \Rightarrow a^3=216
\displaystyle \Rightarrow a^3=6^3
\displaystyle \therefore a=6
\displaystyle \text{The new terms }\frac{a}{r}+2,\ a+8,\ ar+6\text{ are in A.P.}
\displaystyle \therefore 2(a+8)=\left(\frac{a}{r}+2\right)+(ar+6)
\displaystyle \Rightarrow 2(a+8)=\frac{a}{r}+ar+8
\displaystyle \text{Substituting }a=6,\text{ we get}
\displaystyle 2(6+8)=\frac{6}{r}+6r+8
\displaystyle \Rightarrow 28=\frac{6}{r}+6r+8
\displaystyle \Rightarrow \frac{6}{r}+6r=20
\displaystyle \text{Multiplying throughout by }r,
\displaystyle 6+6r^2=20r
\displaystyle \Rightarrow 6r^2-20r+6=0
\displaystyle \Rightarrow 3r^2-10r+3=0
\displaystyle \Rightarrow (3r-1)(r-3)=0
\displaystyle \therefore r=\frac{1}{3}\text{ or }r=3
\displaystyle \text{When }r=3,\text{ the terms are }\frac{6}{3},6,6(3),\text{ i.e. }2,6,18.
\displaystyle \text{When }r=\frac{1}{3},\text{ the terms are }18,6,2,\text{ which are the same in reverse order.}
\displaystyle \therefore \text{The required numbers are }2,6\text{ and }18.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find three numbers in G.P. whose product is }729\text{ and the sum of their products}
\displaystyle \text{taken in pairs is }819.
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the G.P. be }\frac{a}{r},a,ar.
\displaystyle \text{Given }\frac{a}{r}\times a\times ar=729
\displaystyle \Rightarrow a^3=729
\displaystyle \Rightarrow a^3=9^3
\displaystyle \therefore a=9
\displaystyle \text{Also, }\frac{a}{r}\times a+a\times ar+ar\times\frac{a}{r}=819
\displaystyle \Rightarrow \frac{a^2}{r}+a^2r+a^2=819
\displaystyle \Rightarrow \frac{81}{r}+81r+81=819
\displaystyle \Rightarrow \frac{9}{r}+9r+9=91
\displaystyle \text{Multiplying throughout by }r,
\displaystyle 9+9r^2+9r=91r
\displaystyle \Rightarrow 9r^2-82r+9=0
\displaystyle \Rightarrow (r-9)(9r-1)=0
\displaystyle \therefore r=9\text{ or }r=\frac{1}{9}
\displaystyle \text{When }r=9,\text{ the terms are }\frac{9}{9},9,9(9),\text{ i.e. }1,9,81.
\displaystyle \text{When }r=\frac{1}{9},\text{ the terms are }81,9,1,\text{ which are the same in reverse order.}
\displaystyle \therefore \text{The required numbers are }1,9\text{ and }81.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The sum of three numbers in G.P. is }21\text{ and the sum of their squares is }189.
\displaystyle \text{Find the numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required numbers be }a,ar\text{ and }ar^2.
\displaystyle \text{Given }a+ar+ar^2=21
\displaystyle \Rightarrow a(1+r+r^2)=21\qquad\ldots\ldots\ldots\text{(i)}
\displaystyle \text{Also, }a^2+a^2r^2+a^2r^4=189
\displaystyle \Rightarrow a^2(1+r^2+r^4)=189\qquad\ldots\ldots\ldots\text{(ii)}
\displaystyle \text{Squaring both sides of (i), we get}
\displaystyle a^2(1+r+r^2)^2=441
\displaystyle \Rightarrow a^2(1+r^2+r^4)+2a^2r(1+r+r^2)=441
\displaystyle \text{Using (i) and (ii), we get}
\displaystyle 189+2ar(21)=441
\displaystyle \Rightarrow 42ar=252
\displaystyle \therefore ar=6
\displaystyle \Rightarrow a=\frac{6}{r}\qquad\ldots\ldots\ldots\text{(iii)}
\displaystyle \text{Substituting }a=\frac{6}{r}\text{ in (i), we get}
\displaystyle \frac{6}{r}(1+r+r^2)=21
\displaystyle \Rightarrow 6+6r+6r^2=21r
\displaystyle \Rightarrow 6r^2-15r+6=0
\displaystyle \Rightarrow 2r^2-5r+2=0
\displaystyle \Rightarrow (2r-1)(r-2)=0
\displaystyle \therefore r=\frac{1}{2}\text{ or }r=2
\displaystyle \text{When }r=\frac{1}{2},\text{ from }ar=6,\text{ we get }a=12.
\displaystyle \text{Thus, the G.P. is }12,6,3.
\displaystyle \text{When }r=2,\text{ from }ar=6,\text{ we get }a=3.
\displaystyle \text{Thus, the G.P. is }3,6,12.
\displaystyle \therefore \text{The required numbers are }3,6\text{ and }12.
\displaystyle \\

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