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

Answer:

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

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

Also given \frac{a}{r} +  a + ar = 65

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

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

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

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

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

\Rightarrow r = \frac{1}{3}  or r = 3

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

and when a = 15 and r = 3 , G.P. is 5, 15, 45, \ldots

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

Answer:

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

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

Also given \frac{a}{r} +  a + ar = 38

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

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

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

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

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

\Rightarrow r = \frac{2}{3}  or r = \frac{3}{2}

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

and when a = 12 and r = \frac{3}{2} , 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 \frac{13}{12} and their product is -1 . Find the G.P.

Answer:

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

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

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

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

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

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

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

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

\Rightarrow r = \frac{-4}{3}  or r = \frac{-3}{4}

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

and when a = -1 and r = \frac{-3}{4} , 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 125 and the sum of their products taken in pairs is 87 \frac{1}{2} . Find them.

Answer:

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

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

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

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

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

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

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

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

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

\Rightarrow r = \frac{1}{2} \ or \ r = 2

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

and when a = 5 and r = 2 , 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 \frac{39}{10} and their product is 1 . Find the common ratio and the terms.

Answer:

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

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

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

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

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

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

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

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

\Rightarrow r = \frac{5}{2}  or r = \frac{2}{5}

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

and when a = 1 and r = \frac{2}{5} , 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 14 . If the first two terms are each increase by 1 and the third is decreased by 1 , the resulting numbers are in A.P. Find the numbers.

Answer:

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

Therefore \frac{a}{r} + a +  ar = 14

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

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

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

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

Substituting from i) we get

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

\Rightarrow 3a = 12

\Rightarrow a = 4

Substituting the value of a in i) we get

\frac{4}{r} + 4 +  4r = 14

\Rightarrow \frac{2}{r} + 2 +  2r = 7

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

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

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

\Rightarrow r = 2 \ or \ r = \frac{1}{2}

Therefore when a = 4 and r = 2 , G.P. is 8, 4, 2 , \ldots

and when a = 4 and r = \frac{1}{2} , G.P. is 2, 4, 8, \ldots

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

Answer:

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

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

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

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

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

Substituting for a we get

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

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

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

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

\Rightarrow r = 3 \ or \ r = \frac{1}{3}

Therefore when a = 6 and r = 3 , G.P. is 18, 6, 2 , \ldots

and when a = 6 and r = \frac{1}{3} , G.P. is 2, 6, 18, \ldots

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

Answer:

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

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

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

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

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

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

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

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

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

\Rightarrow r = 9 \ or \ r = \frac{1}{9}

Therefore when a = 9 and r = 9 , G.P. is 1, 9, 81 , \ldots

and when a = 9 and r = \frac{1}{9} , G.P. is 81, 9, 1, \ldots

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

Answer:

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

Therefore a + ar + ar^2 = 21

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

Also a^2 + a^2r^2 + a^2 r^4 = 189

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

Squaring both sides of i) we get

a^2 ( 1 + r + r^2)^2  = 441

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

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

\Rightarrow 189 + 2ar  (21)  = 441

\Rightarrow ar = 6

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

Substituting this in i) we get

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

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

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

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

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

\Rightarrow r = \frac{1}{2} \ or \ r = 2

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

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

and when a = 2 and r = 3 , G.P. is 3, 6, 12, \ldots