Question 1: Find three numbers in G.P. whose sum is $\displaystyle 65$ and whose product is $\displaystyle 3375$.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Given } \frac{a}{r} \times a \times ar = 3375 \Rightarrow a^3 = 3375 \Rightarrow a = 15$

$\displaystyle \text{Also given } \frac{a}{r} + a + ar = 65$

$\displaystyle \Rightarrow \frac{15}{r} + 15 + 15r = 65$

$\displaystyle \Rightarrow 15r^2 + 15r + 15 = 65r$

$\displaystyle \Rightarrow 15r^2 - 50r + 15 = 0$

$\displaystyle \Rightarrow 3r^2 - 10r + 3 = 0$

$\displaystyle \Rightarrow ( 3r-1) ( r-3) = 0$

$\displaystyle \Rightarrow r = \frac{1}{3} \text{ or } r = 3$

$\displaystyle \text{Therefore when } a = 15 \text{ and } r = \frac{1}{3} , \text{ G.P. is } 45, 15, 5, \ldots$

$\displaystyle \text{And } \text{When } a = 15 \text{ and } r = 3 , \text{ G.P. is } 5, 15, 45, \ldots$

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Question 2: Find three numbers in G.P. whose sum is $\displaystyle 38$ and whose product is $\displaystyle 1728$.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Given } \frac{a}{r} \times a \times ar = 1728 \Rightarrow a^3 = 1728 \Rightarrow a = 12$

$\displaystyle \text{Also given } \frac{a}{r} + a + ar = 38$

$\displaystyle \Rightarrow \frac{12}{r} + 12 + 12r = 38$

$\displaystyle \Rightarrow 12r^2 + 12r + 12 = 38r$

$\displaystyle \Rightarrow 12r^2 - 26r + 12 = 0$

$\displaystyle \Rightarrow 6r^2 - 13r + 6 = 0$

$\displaystyle \Rightarrow ( 3r-2) ( 2r-3) = 0$

$\displaystyle \Rightarrow r = \frac{2}{3} \text{ or } r = \frac{3}{2}$

$\displaystyle \text{Therefore when } a = 12 \text{ and } r = \frac{2}{3} , \text{ G.P. is } 18, 12, 8 , \ldots$

$\displaystyle \text{And } \text{When } a = 12 \text{ and } r = \frac{3}{2} , \text{ G.P. is } 8, 12, 18, \ldots$

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Question 3: The sum of the first three terms of a G.P. is $\displaystyle \frac{13}{12}$ and their product is $\displaystyle -1$. Find the G.P.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Given } \frac{a}{r} \times a \times ar = -1 \Rightarrow a^3 = -1 \Rightarrow a = -1$

$\displaystyle \text{Also given } \frac{a}{r} + a + ar = \frac{13}{12}$

$\displaystyle \Rightarrow \frac{-1}{r} -1 -1r = \frac{13}{12}$

$\displaystyle \Rightarrow -12r^2 - 12r - 12 = 13r$

$\displaystyle \Rightarrow 12r^2 +25r + 12 = 0$

$\displaystyle \Rightarrow 4r(3r+4)+3(3r+4)=0$

$\displaystyle \Rightarrow (3r+4) ( 4r+3) = 0$

$\displaystyle \Rightarrow r = \frac{-4}{3} \text{ or } r = \frac{-3}{4}$

$\displaystyle \text{Therefore when } a = -1 \text{ and } r = \frac{-4}{3} , \text{ G.P. is } \frac{3}{4} , -1, \frac{4}{3} , \ldots$

$\displaystyle \text{And } \text{When } a = -1 \text{ and } r = \frac{-3}{4} , \text{ G.P. is } \frac{4}{3} , -1, \frac{3}{4} , \ldots$

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Question 4: The product of the first three terms of a G.P. is $\displaystyle 125$ and the sum of their products taken in pairs is $\displaystyle 87 \frac{1}{2}$ . Find them.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Given } \frac{a}{r} \times a \times ar = 125 \Rightarrow a^3 = 125 \Rightarrow a = 5$

$\displaystyle \text{Also } \frac{a}{r} \times a + a \times ar + ar \times \frac{a}{r} = \frac{175}{2}$

$\displaystyle \Rightarrow \frac{a^2}{r} + a^2 r + a^2 = \frac{175}{2}$

$\displaystyle \Rightarrow \frac{25}{r} + 25 r + 25 = \frac{175}{2}$

$\displaystyle \Rightarrow \frac{1}{r} + r + 1 = \frac{7}{2}$

$\displaystyle \Rightarrow 2r^2 + 2r + 2 = 7r$

$\displaystyle \Rightarrow 2r^2 -5r + 2 = 0$

$\displaystyle \Rightarrow (2r-1)(r-2) = 0$

$\displaystyle \Rightarrow r = \frac{1}{2} \text{ or } r = 2$

$\displaystyle \text{Therefore when } a = 5 \text{ and } r = \frac{1}{2} , \text{ G.P. is } 10, 5, \frac{5}{2} , \ldots$

$\displaystyle \text{And } \text{When } a = 5 \text{ and } r = 2 , \text{ G.P. is } \frac{5}{2} , 5, 10, \ldots$

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Question 5: The sum of the first three terms of a G.P. is $\displaystyle \frac{39}{10}$ and their product is $\displaystyle 1$. Find the common ratio and the terms.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Given } \frac{a}{r} \times a \times ar = 1 \Rightarrow a^3 = 1 \Rightarrow a = 1$

$\displaystyle \text{Also given } \frac{a}{r} + a + ar = \frac{39}{10}$

$\displaystyle \Rightarrow \frac{1}{r} +1 + r = \frac{39}{10}$

$\displaystyle \Rightarrow 10r^2 +10r + 10 = 39r$

$\displaystyle \Rightarrow 10r^2-29r+10 = 0$

$\displaystyle \Rightarrow 5r(2r-5)-2 (2r-5)=0$

$\displaystyle \Rightarrow (2r-5) ( 5r-2) = 0$

$\displaystyle \Rightarrow r = \frac{5}{2} \text{ or } r = \frac{2}{5}$

$\displaystyle \text{Therefore when } a = 1 \text{ and } r = \frac{5}{2} , \text{ G.P. is } \frac{5}{2} , 1, \frac{2}{5} , \ldots$

$\displaystyle \text{And } \text{When } a = 1 \text{ and } r = \frac{2}{5} , \text{ G.P. is } \frac{2}{5} , 1, \frac{5}{2} , \ldots$

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Question 6: The sum of three numbers in G.P. is $\displaystyle 14$. If the first two terms are each increase by $\displaystyle 1$ and the third is decreased by $\displaystyle 1$, the resulting numbers are in A.P. Find the numbers.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Therefore } \frac{a}{r} + a + ar = 14$

$\displaystyle \Rightarrow a ( \frac{1}{r} +1+r) = 14$ … … … … … i)

$\displaystyle \text{Also, new terms } \frac{a}{r} +1 , a+1 , ar - 1 \text{ are in A.P. }$

$\displaystyle \therefore 2 ( a+1) = ( \frac{a}{r} +1) + (ar - 1)$

$\displaystyle \Rightarrow 2 ( a+1) = \frac{a}{r} + ar$

Substituting from i) we get

$\displaystyle \Rightarrow 2 ( a+1) = 14 - a$

$\displaystyle \Rightarrow 3a = 12$

$\displaystyle \Rightarrow a = 4$

Substituting the value of a in i) we get

$\displaystyle \frac{4}{r} + 4 + 4r = 14$

$\displaystyle \Rightarrow \frac{2}{r} + 2 + 2r = 7$

$\displaystyle \Rightarrow 2r^2+ 2r + 2 = 7r$

$\displaystyle \Rightarrow 2r^2 - 5r+2 = 0$

$\displaystyle \Rightarrow (r-2)(2r-1) = 0$

$\displaystyle \Rightarrow r = 2 \text{ or } r = \frac{1}{2}$

$\displaystyle \text{Therefore when } a = 4 \text{ and } r = 2 , \text{ G.P. is } 8, 4, 2 , \ldots$

$\displaystyle \text{And } \text{When } a = 4 \text{ and } r = \frac{1}{2} , \text{ G.P. is } 2, 4, 8, \ldots$

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Question 7: The product of three numbers in G.P. is $\displaystyle 216$. If $\displaystyle 2, 8, 6$ be added to them, the results are in A.P.. Find the numbers.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Therefore } \frac{a}{r} \times a \times ar = 216 \Rightarrow a^3 = 216 \Rightarrow a = 6$

$\displaystyle \text{Also, new terms } \frac{a}{r} +2 , a+8 , ar +6 \text{ are in A.P. }$

$\displaystyle \therefore 2 ( a+8) = ( \frac{a}{r} +2) + (ar +6)$

$\displaystyle \Rightarrow 2 ( a+8) = \frac{a}{r} + ar + 8$

Substituting for a we get

$\displaystyle \Rightarrow 2 ( 6+8) = \frac{6}{r} + 6r + 8$

$\displaystyle \Rightarrow 20 = \frac{6}{r} +6r$

$\displaystyle \Rightarrow 6r^2 -20r + 6 = 0$

$\displaystyle \Rightarrow (r-3)(6r-2) = 0$

$\displaystyle \Rightarrow r = 3 \text{ or } r = \frac{1}{3}$

$\displaystyle \text{Therefore when } a = 6 \text{ and } r = 3 , \text{ G.P. is } 18, 6, 2 , \ldots$

$\displaystyle \text{And } \text{When } a = 6 \text{ and } r = \frac{1}{3} , \text{ G.P. is } 2, 6, 18, \ldots$

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Question 8: Find three numbers in G.P. whose product is $\displaystyle 729$ and the sum of their product in pairs is $\displaystyle 819$.

$\displaystyle \text{Let the terms of the G.P. be } \frac{a}{r} , a, ar$

$\displaystyle \text{Therefore } \frac{a}{r} \times a \times ar = 729 \Rightarrow a^3 = 729 \Rightarrow a = 9$

$\displaystyle \text{Also } ( \frac{a}{r} \times a ) + ( a \times ar) + ( ar \times \frac{a}{r} ) = 819$

$\displaystyle \Rightarrow ( \frac{9}{r} \times 9 ) + ( 9 \times 9r) + ( 9r \times \frac{9}{r} ) = 819$

$\displaystyle \Rightarrow \frac{81}{r} + 81r + 81 = 819$

$\displaystyle \Rightarrow \frac{9}{r} + 9r + 9 = 91$

$\displaystyle \Rightarrow 9r^2 + 9r + 9 = 91 r$

$\displaystyle \Rightarrow 9r^2 - 82r+9 = 0$

$\displaystyle \Rightarrow ( r-9) (9r-1) = 0$

$\displaystyle \Rightarrow r = 9 \text{ or } r = \frac{1}{9}$

$\displaystyle \text{Therefore when } a = 9 \text{ and } r = 9 , \text{ G.P. is } 1, 9, 81 , \ldots$

$\displaystyle \text{And } \text{When } a = 9 \text{ and } r = \frac{1}{9} , \text{ G.P. is } 81, 9, 1, \ldots$

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Question 9: The sum of three numbers in G.P. is $\displaystyle 21$ and the sum of their squares is $\displaystyle 189$. Find the numbers.

$\displaystyle \text{Let the required numbers be } a, ar \text{ and } ar^2$

$\displaystyle \text{Therefore } a + ar + ar^2 = 21$

$\displaystyle \Rightarrow a ( 1 + r + r^2) = 21$ … … … … … i)

$\displaystyle \text{Also } a^2 + a^2r^2 + a^2 r^4 = 189$

$\displaystyle \Rightarrow a^2 ( 1 + r^2 + r^4) = 189$ … … … … … ii)

Squaring both sides of i) we get

$\displaystyle a^2 ( 1 + r + r^2)^2 = 441$

$\displaystyle \Rightarrow a^2 ( 1 + r^2 + r^4) + 2a^2r ( 1 + r + r^2)= 441$

$\displaystyle \Rightarrow 189+ 2ar [ a(1 + r + r^2)] = 441$

$\displaystyle \Rightarrow 189 + 2ar (21) = 441$

$\displaystyle \Rightarrow ar = 6$

$\displaystyle \Rightarrow a = \frac{6}{r}$ … … … … … iii)

Substituting this in i) we get

$\displaystyle \frac{6}{r} ( 1 + r + r^2) = 21$

$\displaystyle \Rightarrow 6r^2 + 6r + 6 = 21 r$

$\displaystyle \Rightarrow 6r^2 - 15r + 6 = 0$

$\displaystyle \Rightarrow 2r^2 - 5r + 2 = 0$

$\displaystyle \Rightarrow (2r-1)(r-2) = 0$

$\displaystyle \Rightarrow r = \frac{1}{2} \text{ or } r = 2$

$\displaystyle \text{When } r = \frac{1}{2} , a = 12 \text{ and when } r = 2, a = 3$

$\displaystyle \text{Therefore when } a = 12 \text{ and } r = \frac{1}{2} , \text{ G.P. is } 12, 6, 3 , \ldots$

$\displaystyle \text{And } \text{When } a = 2 \text{ and } r = 3 , \text{ G.P. is } 3, 6, 12, \ldots$