Question 1: Show that each one of the following progressions is a G.P.. Also, find the common ratio in each case:

\displaystyle \text{i) } 4, -2, 1, - \frac{1}{2} , \ldots \hspace{1.0cm} \text{ii) } \frac{-2}{3} , -6, -54, \ldots \hspace{1.0cm} \text{iii) } a, \frac{3a^2}{4} , \frac{9a^3}{16} , \ldots \hspace{1.0cm} \text{iv) } \frac{1}{2} , \frac{1}{3} , \frac{2}{9} , \frac{4}{27} , \ldots  

Answer:

\displaystyle \text{i) } 4, -2, 1, - \frac{1}{2} , \ldots

\displaystyle \text{Given } a_1 = 4, \ \ \ \ a_2 = -2, \ \ \ \ a_3 = 1, \ \ \ \ a_4 = \frac{-1}{2}  

\displaystyle \therefore \frac{a_2}{a_1} = \frac{-2}{4} = \frac{-1}{2} \text{ and } \frac{a_3}{a_2} = \frac{-1}{2} \text{ and } \frac{a_4}{a_3} = \frac{\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{Therefore } a_1, a_2, a_3 \text{ and } a_4 \text{ are in G.P. } \text{where } a = 4 \text{ and } r = \frac{-1}{2}  

\displaystyle \text{ii) } \frac{-2}{3} , -6, -54, \ldots

\displaystyle \text{Given } a_1 = \frac{-2}{3} \ \ \ \ a_2 = -6, \ \ \ \ a_3 = -54

\displaystyle \therefore \frac{a_2}{a_1} = \frac{-6}{\frac{-2}{3}} = 9 \frac{a_3}{a_2} = \frac{-54}{-6}  

\displaystyle \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = 9

\displaystyle \text{Therefore } a_1, a_2, a_3 \text{ are in G.P. } \text{where } a = \frac{-2}{3} \text{ and } r = 9

\displaystyle \text{iii) } a, \frac{3a^2}{4} , \frac{9a^3}{16} , \ldots

\displaystyle \text{Given } a_1 = a a_2 = \frac{3a^2}{4} , \ \ \ \ a_3 = \frac{9a^3}{16}  

\displaystyle \therefore \frac{a_2}{a_1} = \frac{3a^2}{4a} = \frac{3a}{4} \text{ and } \frac{a_3}{a_2} = \frac{9a^3}{16} \cdot \frac{4}{3a^2} = \frac{3a}{4}  

\displaystyle \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{3a}{4}  

\displaystyle \text{Therefore } a_1, a_2, a_3 \text{ are in G.P. } \text{where } a \text{ is the first term and } r = \frac{3a}{4}  

\displaystyle \text{iv) } \frac{1}{2} , \frac{1}{3} , \frac{2}{9} , \frac{4}{27} , \ldots

\displaystyle \text{Given } a_1 = \frac{1}{2} , \ \ \ \ a_2 = \frac{1}{3} , \ \ \ \ a_3 = \frac{2}{9} , \ \ \ \ a_4 = \frac{4}{27}  

\displaystyle \therefore \frac{a_2}{a_1} = \frac{1/3}{1/2} = \frac{2}{3} \text{ and } \frac{a_3}{a_2} = \frac{2/9}{1/3} = \frac{2}{3} \text{ and } \frac{a_4}{a_3} = \frac{4/27}{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{Therefore } a_1, a_2, a_3 \text{ and } a_4 \text{ are in G.P. } \text{where } a = \frac{1}{2} \text{ and } r = \frac{2}{3}  

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\displaystyle \text{Question 2: Show that the sequence defined by } a_n = \frac{2}{3^n} , n \in N \text{ is a G.P. }

Answer:

\displaystyle \text{Given } a_n = \frac{2}{3^n} , n \in N is a G.P

\displaystyle \text{Given } a_1 = \frac{2}{3} , \ \ \ \ a_2 = \frac{2}{3^2} , \ \ \ \ a_3 = \frac{2}{3^3} , \ \ \ \ a_4 = \frac{2}{3^4}  

\displaystyle \therefore \frac{a_2}{a_1} = \frac{2/3^2}{2/3} = \frac{1}{3} \text{ and } \frac{a_3}{a_2} = \frac{2/3^3}{2/3^2} = \frac{1}{3} \text{ and } \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{Therefore } a_n = \frac{2}{3^n} , n \in N \text{is a G.P where } a = \frac{2}{3} \text{ and } r = \frac{1}{3}  

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Question 3: Find:

\displaystyle \text{i) the } 9^{th} \text{ term of the G.P. } 1, 4, 16, 64 \ldots

\displaystyle \text{ii) the } {10}^{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^{th} \text{ term of the G.P. } 0.3, 0.06, 0.012, \ldots

\displaystyle \text{iv) the } {12}^{th} \text{ term of the G.P. } \frac{1}{a^3x^3} , ax, a^5x^5, \ldots

\displaystyle \text{v) the } n^{th} \text{ term of the G.P. } \sqrt{3}, \frac{1}{\sqrt{3}} , \frac{1}{3\sqrt{3}} , \ldots

\displaystyle \text{vi) the } {10} ^{th} \text{ term of the G.P. } \sqrt{2}, \frac{1}{\sqrt{2}} , \frac{1}{2\sqrt{2}} \ldots

Answer:

\displaystyle \text{i) To find the } 9^{th} \text{ term of the G.P. } 1, 4, 16, 64 \ldots

\displaystyle \text{Here } a = 1, \hspace{1.0cm} r = \frac{a_2}{a_1} = \frac{4}{1} = 4

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_9 = (1)(4)^{9-1} = 4^8 = 65536

\displaystyle \text{ii) To find the } {10}^{th} \text{term of the G.P.} \frac{-3}{4} , \frac{1}{2} , \frac{-1}{3} , \frac{2}{9} \ldots

\displaystyle \text{Here } a = \frac{-3}{4} , \hspace{1.0cm} r = \frac{a_2}{a_1} = \Big( \frac{1/2}{-3/4} \Big) = \frac{-2}{3}  

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_{10} = \Big( \frac{-3}{4} \Big) \Big( \frac{-2}{3} \Big)^{10-1} = \frac{1}{2} \Big ( \frac{2}{3} \Big )^8

\displaystyle \text{iii) To find the } 8^{th} \text{ term of the G.P. } 0.3, 0.06, 0.012, \ldots

\displaystyle \text{Here } a = 0.3, \hspace{1.0cm} r = \frac{a_2}{a_1} = \frac{0.06}{0.3} = 0.2

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_8 = (0.3)(0.2)^{8-1} = (0.3)(0.2)^7

\displaystyle \text{iv) To find the } {12}^{th} \text{term of the G.P.} \frac{1}{a^3x^3} , ax, a^5x^5, \ldots

\displaystyle \text{Here } a = \frac{1}{a^3x^3} , \hspace{1.0cm} r = \frac{a_2}{a_1} = \Bigg( \frac{ax}{\frac{1}{a^3x^3}} \Bigg) = a^4x^4

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_{12} = \Big( \frac{1}{a^3x^3} \Big) (a^4 x^4)^{12-1} = a^{41}x^{41}

\displaystyle \text{v) To find the } n^{th} \text{term of the G.P.} \sqrt{3}, \frac{1}{\sqrt{3}} , \frac{1}{3\sqrt{3}} , \ldots

\displaystyle \text{Here } a = \sqrt{3}, \hspace{1.0cm} r = \frac{a_2}{a_1} = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3}  

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_n = (\sqrt{3}) \Big( \frac{1}{3} \Big)^{n-1}

\displaystyle \text{vi) To find the } {10} ^{th} \text{term of the G.P.} \sqrt{2}, \frac{1}{\sqrt{2}} , \frac{1}{2\sqrt{2}} \ldots

\displaystyle \text{Here } a = \sqrt{2}, \hspace{1.0cm} r = \frac{a_2}{a_1} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}  

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_{10} = (\sqrt{2}) \Big( \frac{1}{2} \Big)^{10-1} = \sqrt{2} \Big( \frac{1}{2^9} \Big) = \frac{1}{(\sqrt{2})^{17}}  

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\displaystyle \text{Question 4: Find the 4th term from the end of the G.P. } \frac{2}{27} , \frac{2}{9}, \frac{2}{3} , \ldots , 162

Answer:

\displaystyle \text{Here } a = \frac{2}{27} , \hspace{1.0cm} r = \frac{a_2}{a_1} = \Big( \frac{2/9}{2/27} \Big) = 3

\displaystyle \text{Last term }  l = 162

Now, when we reverse the G.P., we have \displaystyle a = l \text{ and }  r = \frac{1}{3}  

We know, \displaystyle a_n = ar^{n-1}

\displaystyle \therefore a_{4} = (162) ( \frac{1}{3} ) = \frac{162}{3^3} = 6

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Question 5: Which term of the progression \displaystyle 0.004, 0.02, 0.1, \ldots \text{ is } 12.5

Answer:

\displaystyle \text{Given series: } 0.004, 0.02, 0.1, \ldots is \displaystyle 12.5

\displaystyle \text{Here } a = 0.004 \frac{a_2}{a_1} = \frac{0.02}{0.004} = 5 \frac{a_3}{a_2} = \frac{0.1}{0.02} = 5

\displaystyle \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = 5

\displaystyle \text{Let }  12.5 \text{ be the }  n^{th} term

\displaystyle \therefore 12.5 = ar^{n-1}

\displaystyle \Rightarrow 12.5 = ( 0.004) ( 5)^{n-1}

\displaystyle \Rightarrow 5^{n-1} = 3125

\displaystyle \Rightarrow 5^{n-1} = 5^5

\displaystyle \therefore n - 1 = 5 \Rightarrow n = 6

\displaystyle \text{Therefore } 12.5 \text{ is the } 6^{th} \text{ term of the given G.P. }

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Question 6: Which term of the G.P.:

\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} ?

Answer:

\displaystyle \text{i) Given series: } \sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{4\sqrt{2}} , \ldots ?

\displaystyle \text{Here } a = \sqrt{2} \frac{a_2}{a_1} = \frac{1/\sqrt{2}}{\sqrt{2}} = \frac{1}{2}  

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \text{Let }  \frac{1}{512\sqrt{2}} \text{ be the }  n^{th} term

\displaystyle \therefore \ \ \ \frac{1}{512\sqrt{2}} = ( \sqrt{2}) \Big( \frac{1}{2} \Big)^{n-1}

\displaystyle \Rightarrow \frac{1}{1024} = \Big( \frac{1}{2} \Big)^{n-1}

\displaystyle \Rightarrow \Big( \frac{1}{2} \Big)^{10} = \Big( \frac{1}{2} \Big)^{n-1}

\displaystyle \Rightarrow n-1 = 10

\displaystyle \Rightarrow n = 11

\displaystyle \text{Therefore } \frac{1}{512\sqrt{2}} \text{ is the } {11}^{th} \text{ term of the given G.P. }

\displaystyle \text{ii) Given series: } 2, 2\sqrt{2}, 4, \ldots

\displaystyle \text{Here } a = 2 \frac{a_2}{a_1} = \frac{2\sqrt{2}}{2} = \sqrt{2}

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \text{Let }  128 \text{ be the }  n^{th} term

\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

\displaystyle \Rightarrow n = 13

\displaystyle \text{Therefore } 128 \text{ is the } {13}^{th} \text{ term of the given G.P. }

\displaystyle \text{iii) Given series: } \sqrt{3}, 3, 3\sqrt{3}, \ldots

\displaystyle \text{Here } a = \sqrt{3} \frac{a_2}{a_1} = \frac{3}{\sqrt{3}} = \sqrt{3}

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \text{Let }  729 \text{ be the }  n^{th} term

\displaystyle \therefore 729 = (\sqrt{3}) ( \sqrt{3} )^{n-1}

\displaystyle \Rightarrow (\sqrt{3})^{n-1} = \frac{( \sqrt{2})^{12}}{\sqrt{3}}  

\displaystyle \Rightarrow (\sqrt{3})^{n-1} = ( \sqrt{3})^{11}

\displaystyle \Rightarrow n - 1 = 11

\displaystyle \Rightarrow n = 12

\displaystyle \text{Therefore } 729 \text{ is the } {12}^{th} \text{ term of the given G.P. }

\displaystyle \text{iv) Given series: } \frac{1}{3}, \frac{1}{9}, \frac{1}{27} , \ldots

\displaystyle \text{Here } a = \frac{1}{3} \frac{a_2}{a_1} = \frac{1/9}{1/3} = \frac{1}{3}  

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \text{Let }  \frac{1}{19683} \text{ be the }  n^{th} term

\displaystyle \therefore \frac{1}{19683} = \Big( \frac{1}{3} \Big) \Big( \frac{1}{3} \Big )^{n-1}

\displaystyle \Rightarrow (\sqrt{3})^{n-1} = \frac{( \sqrt{2})^{12}}{\sqrt{3}}  

\displaystyle \Rightarrow (\frac{1}{3})^{9} = \frac{( \sqrt{2})^{12}}{\sqrt{3}}  

\displaystyle \Rightarrow n - 1 + 1 = 9

\displaystyle \Rightarrow n = 9

\displaystyle \text{Therefore } \frac{1}{19683} \text{ is the } {9}^{th} \text{ term of the given G.P. }

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Question 7: Which term of the progression \displaystyle 18, -12, 8, \ldots \text{ is }  \frac{512}{729} ?

Answer:

\displaystyle \text{Given series: } 18, -12, 8, \ldots

\displaystyle \text{Here } a = 18 \frac{a_2}{a_1} = \frac{-12}{18} = \frac{-2}{3}  

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \text{Let }  \frac{512}{729} \text{ be the }  n^{th} term

\displaystyle \Rightarrow \frac{512}{729} = ( 18) \Big( \frac{-2}{3} \Big)^{n-1}

\displaystyle \Rightarrow \Big( \frac{-2}{3} \Big)^{n-1} = \frac{512}{729} \times \frac{1}{18}  

\displaystyle \Rightarrow \Big( \frac{-2}{3} \Big)^{n-1} = \frac{256}{6561}  

\displaystyle \Rightarrow \Big( \frac{-2}{3} \Big)^{n-1} = \Big( \frac{-2}{3} \Big)^{8}

\displaystyle \therefore n - 1 = 8 \Rightarrow n = 9

\displaystyle \text{Therefore } \frac{512}{729} \text{ is the } 9^{th} \text{ term of the given G.P. }

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\displaystyle \text{Question 8: Find the 4th terms from the end of the G.P. } \frac{1}{2}, \frac{1}{6}, \frac{1}{8}, \frac{1}{54} , \ldots \frac{1}{4374}  

Answer:

\displaystyle \text{Given series: } \frac{1}{2}, \frac{1}{6}, \frac{1}{8}, \frac{1}{54} , \ldots

\displaystyle \text{Reversing the G.P. we get } a = \frac{1}{4374} , \hspace{1.0cm} r = 3

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \therefore a_4 = \Big( \frac{1}{4374} \Big) (3)^{4-1}

\displaystyle \Rightarrow a_4 = \Big( \frac{1}{4374} \Big) (3)^{3}

\displaystyle \Rightarrow a_4 = \frac{1}{162}  

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\displaystyle \text{Question 9: The 4th term of the G.P. is } 27 \text{and the } 7^{th} \text{ term is } 729 , \text{ find the G.P. }

Answer:

\displaystyle \text{Given } a_4 = 27 \Rightarrow ar^3 = 27 … … … … … i)

\displaystyle \text{Given } a_7 = 729 \Rightarrow ar^6 = 729 … … … … … ii)

Dividing ii) by i) we get

\displaystyle \frac{ar^6}{ar^3} = \frac{729}{27}  

\displaystyle \Rightarrow r^3 = 27

\displaystyle \Rightarrow r^3 = 3^3

\displaystyle \Rightarrow r = 3

Substituting in i) we get

\displaystyle a(3)^3 = 27 \Rightarrow a = 1

\displaystyle \text{Therefore the G.P. is } 1, 3, 9, 27, \ldots

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Question 10:The \displaystyle 7^{th} term of the G.P. is \displaystyle 8 times the \displaystyle 4^{th} term and \displaystyle 5^{th} term is \displaystyle 48 . Find the G.P.

Answer:

\displaystyle \text{Given } a_7 = a_4

\displaystyle \Rightarrow ar^6 = 8 ar^3

\displaystyle \Rightarrow r^3 = 8

\displaystyle \Rightarrow r = 2

\displaystyle \text{Substituting we get } a(2)^4 = 48 \Rightarrow a = \frac{48}{4 \times 14} =3

\displaystyle \text{Therefore the G.P. is } 3, 6, 12, 24, \ldots

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Question 11: If the G.P.’s \displaystyle 5, 10, 20, \ldots \text{ and } 1280, 640, 320, \ldots have their \displaystyle n^{th} terms equal, find the value of \displaystyle n .

Answer:

\displaystyle \text{For G.P. } 5, 10, 20, \ldots

\displaystyle \text{Therefore } a = 5 \frac{a_2}{a_1} = \frac{10}{5} = 2

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \Rightarrow a_n = (5)( 2)^{n-1} … … … … … i)

\displaystyle \text{For G.P. } 1280, 640, 320, \ldots

\displaystyle \text{Therefore } a = 1280 \frac{a_2}{a_1} = \frac{640}{1280} = \frac{1}{2}  

\displaystyle \text{We know } a_n = ar^{n-1}

\displaystyle \Rightarrow a_n = (1280) \Big( \frac{1}{2} \Big)^{n-1} … … … … … ii)

From i) and ii) we get

\displaystyle (5)( 2)^{n-1} = (1280) \Big( \frac{1}{2} \Big)^{n-1}

\displaystyle \Rightarrow 2^{2n - 2} = 256

\displaystyle \Rightarrow 2^{2n - 2} = 2^8

\displaystyle \Rightarrow 2n - 2 = 8

\displaystyle \Rightarrow n = 5

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Question 12: The \displaystyle 5^{th}, 8^{th} \text{ and } {11}^{th} term of a G.P. are \displaystyle p, q \text{ and }  s respectively, prove tht \displaystyle q^2 = ps .

Answer:

\displaystyle \text{Given } a_5 = ar^4 = p … … … … … i)

\displaystyle a_8 = ar^7 = q … … … … … ii)

\displaystyle a_{11} = ar^{10} = s … … … … … iii)

\displaystyle \text{Therefore } ps = (ar^4) \cdot ( ar^{10} ) = ( ar^7)^{2} = q^2 . Hence proved.

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Question 13: The \displaystyle 4^{th} term of a G.P. is square of its \displaystyle 2^{nd} term and the first term is \displaystyle -3 . Find its \displaystyle 7^{th} term.

Answer:

\displaystyle \text{Given } a_4 = (a_2)^2 \text{ and } a = -3

\displaystyle \Rightarrow ar^3 = ( ar)^2

\displaystyle \Rightarrow ar^3 = a^2 r^2

\displaystyle \Rightarrow r = a

\displaystyle \therefore a_7 = ar^6 = ( -3)( -3)^6 = (-3)^7 = - 2187

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Question 14: In a G.P. the \displaystyle 3^{rd} term is \displaystyle 24 and the \displaystyle 6^{th} term is \displaystyle 192 . Find the \displaystyle {10}^{th} term.

Answer:

\displaystyle ar^2 = 24 … … … … … i)

\displaystyle ar^5 = 192 … … … … … ii)

Dividing ii) and i) we get

\displaystyle r^3 = \frac{192}{24} = 8

\displaystyle r^3 = 2^3 \Rightarrow r = 2

Substituting in i) we get

\displaystyle a(2)^2 = 24 \Rightarrow a = 6

\displaystyle \therefore a_{10} = ar^9 = 6(2)^9 = 3072

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Question 15: If \displaystyle a, b, c \text{ and } p are different real numbers such that

\displaystyle (a^2 + b^2 + c^2 )p^2 - 2 ( ab + bc + cd) p + ( b^2 + c^2 +d^2) \leq 0 , then show that \displaystyle a, b, c \text{ and } d are in G.P.

Answer:

\displaystyle \text{Given } (a^2 + b^2 + c^2 )p^2 - 2 ( ab + bc + cd) p + ( b^2 + c^2 +d^2) \leq 0

\displaystyle \Rightarrow (a^2p^2 + b^2p^2 + c^2p^2 ) - 2 ( abp + bcp + cdp)+ ( b^2 + c^2 +d^2) \leq 0

\displaystyle \Rightarrow (a^2p^2 - 2abp+b^2) + ( b^2p^2 - 2bcp +c^2) + ( c^2 p^2 - 2cdp + d^2) \leq 0

\displaystyle \Rightarrow (ap-b)^2 + ( bp-c)^2 + ( cp-d)^2 \leq 0

Since all the terms are square, therefore they cannot be less than zero. Hence,

\displaystyle \Rightarrow (ap-b)^2 + ( bp-c)^2 + ( cp-d)^2 = 0

\displaystyle \Rightarrow (ap-b)^2 = 0 \hspace{1.5cm} ( bp-c)^2=0 \hspace{1.5cm} ( cp-d)^2 = 0

\displaystyle \Rightarrow p = \frac{b}{a} p = \frac{c}{b} p = \frac{d}{c}  

\displaystyle \Rightarrow \frac{b}{a} = \frac{c}{b} = \frac{d}{c}  

\displaystyle \Rightarrow a, b, c \text{ and } d are in G.P.

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\displaystyle \text{Question 16: If } \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} =\frac{c+dx}{c-dx} (x \neq 0) , \text{ then show that } a, b, c \text{ and } d \text{ are in G.P. }

Answer:

\displaystyle \text{Given } \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx} =\frac{c+dx}{c-dx}  

Take the first two terms

\displaystyle \frac{a+bx}{a-bx} = \frac{b+cx}{b-cx}  

Apply componendo and dividendo we get

\displaystyle \frac{(a+bx)+ (a-bx)}{(a+bx) - (a-bx)} = \frac{(b+cx) + (b-cx)}{(b+cx)+ (b-cx)}  

\displaystyle \frac{2a}{2bx} = \frac{2b}{2cx}  

\displaystyle \frac{a}{b} = \frac{b}{c} … … … … … i)

Similarly, Take the last two terms

\displaystyle \frac{b+cx}{b-cx} =\frac{c+dx}{c-dx}  

Apply componendo and dividendo we get

\displaystyle \frac{(b+cx) + (b-cx)}{(b+cx)+ (b-cx)} = \frac{(c+dx)+ (c-dx)}{(c+dx) - (c-dx)}  

\displaystyle \frac{2b}{2cx} = \frac{2c}{2dx}  

\displaystyle \frac{b}{c} = \frac{c}{d} … … … … … ii)

\displaystyle \therefore \frac{a}{b} = \frac{b}{c}= \frac{c}{d}  

\displaystyle \text{Therefore } a, b, c \text{ and } d are in G.P.

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Question 17: If the \displaystyle p^{th} and the \displaystyle q^{th} terms of a G.P. are \displaystyle q \text{ and } p respectively, show that \displaystyle (p+q)^{th} term is \displaystyle \Big( \frac{q^p}{p^q} \Big)^{\frac{1}{p-q}}  

Answer:

\displaystyle \text{Given } a_p = ar^{p-1} = q … … … … … i)

\displaystyle a_q = ar^{q-1} = p … … … … … ii)

Dividing i) by ii)

\displaystyle \frac{ar^{p-1}}{ar^{q-1}} = \frac{q}{p}  

\displaystyle \frac{r^{p-1}}{r^{q-1}} = \frac{q}{p}  

\displaystyle r^{p-1-q+1} = \frac{q}{p}  

\displaystyle r^{p-q} = \frac{q}{p}  

\displaystyle r = \Big( \frac{q}{p} \Big)^{\frac{1}{p-q}} … … … … … i)

Substituting the value of r in ii) we get

\displaystyle a \Big[ \Big( \frac{q}{p} \Big)^{\frac{1}{p-q}} \Big] ^{q-1} = p  

\displaystyle a \Big[ \Big( \frac{q}{p} \Big)^{\frac{q-1}{p-q}} \Big] = p  

\displaystyle a = p \Big( \frac{p}{q} \Big)^{\frac{q-1}{p-q}} … … … … … ii)

\displaystyle \text{Therefore } a_{p+q} = a r^{p+q-1} … … … … … iii)

Substituting i) and ii) in iii) we get

\displaystyle a_{p+q} = \Big[ p \Big( \frac{p}{q} \Big)^{\frac{q-1}{p-q}} \Big] \cdot \Big[ \Big( \frac{q}{p} \Big)^{\frac{1}{p-q}} \Big]^{p+q-1}  

\displaystyle = \Big[ p \Big( \frac{p}{q} \Big)^{\frac{q-1}{p-q}} \Big] \cdot \Big[ \Big( \frac{q}{p} \Big)^{\frac{p+q-1}{p-q}} \Big]  

\displaystyle = \Big[ p \Big( \frac{q}{p} \Big)^{\frac{-(q-1)}{p-q}} \Big] \cdot \Big[ \Big( \frac{q}{p} \Big)^{\frac{p+q-1}{p-q}} \Big]  

\displaystyle = p \Big( \frac{q}{p} \Big)^{ \frac{p+q-1}{p-q} - \frac{q-1}{p-q} }  

\displaystyle = p \Big( \frac{q}{p} \Big)^{ \frac{p}{p-q} }  

\displaystyle = \frac{ q^{\frac{p}{p-q}} }{ p^{\frac{q}{p-q} } }  

\displaystyle = \Big( \frac{q^p}{p^q} \Big)^{\frac{1}{p-q}}