Class 11: Geometric Progression – Exercise 20.1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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:

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

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

$\displaystyle \therefore \frac{a_2}{a_1} = \frac{-2}{4} = \frac{-1}{2} \text{ and } \frac{a_3}{a_2} = \frac{-1}{2} \text{ and } \frac{a_4}{a_3} = \frac{\frac{-1}{2}}{1} = \frac{-1}{2}$

$\displaystyle \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{-1}{2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \therefore \frac{a_2}{a_1} = \frac{1/3}{1/2} = \frac{2}{3} \text{ and } \frac{a_3}{a_2} = \frac{2/9}{1/3} = \frac{2}{3} \text{ and } \frac{a_4}{a_3} = \frac{4/27}{2/9} = \frac{2}{3}$

$\displaystyle \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{2}{3}$

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

$\displaystyle \\$

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

Answer:

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

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

$\displaystyle \therefore \frac{a_2}{a_1} = \frac{2/3^2}{2/3} = \frac{1}{3} \text{ and } \frac{a_3}{a_2} = \frac{2/3^3}{2/3^2} = \frac{1}{3} \text{ and } \frac{a_4}{a_3} = \frac{2/3^4}{2/3^3} = \frac{1}{3}$

$\displaystyle \therefore \frac{a_2}{a_1} = \frac{a_3}{a_2} = \frac{a_4}{a_3} = \frac{1}{3}$

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

$\displaystyle \\$

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow n - 1 + 1 = 9$

$\displaystyle \Rightarrow n = 9$

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

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \\$

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

Answer:

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \\$

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:

$\displaystyle \text{Given } a_7 = a_4$

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

$\displaystyle \Rightarrow r^3 = 8$

$\displaystyle \Rightarrow r = 2$

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow 2n - 2 = 8$

$\displaystyle \Rightarrow n = 5$

$\displaystyle \\$

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.

$\displaystyle \\$

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:

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

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

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

$\displaystyle \Rightarrow r = a$

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

$\displaystyle \\$

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

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

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

$\displaystyle \\$

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.

$\displaystyle \\$

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

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.

$\displaystyle \\$

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