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

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

Answer:

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

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

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

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

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

ii)    \frac{-2}{3} , -6, -54, \ldots

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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Question 3: Find:

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

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

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

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

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

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

Answer:

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

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

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

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

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

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, a_n = ar^{n-1}

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

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

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

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

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

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

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, a_n = ar^{n-1}

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

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

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

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

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

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

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

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

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

Answer:

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

Last term l = 162

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

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

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

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

Answer:

Given series: 0.004, 0.02, 0.1, \ldots is 12.5

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

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

Let 12.5 be the n^{th} term

\therefore  12.5 = ar^{n-1}

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

\Rightarrow 5^{n-1} = 3125

\Rightarrow 5^{n-1} = 5^5

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

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

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

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}} ?

ii) 2, 2\sqrt{2}, 4, \ldots \text{ is } 128 ?

iii) \sqrt{3}, 3, 3\sqrt{3}, \ldots \text{ is } 729?

iv) \frac{1}{3}, \frac{1}{9}, \frac{1}{27} , \ldots \text{ is } \frac{1}{19683} ?

Answer:

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

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

We know a_n = ar^{n-1}

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

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

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

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

\Rightarrow  n-1  =  10 

\Rightarrow  n  =  11 

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

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

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

We know a_n = ar^{n-1}

Let 128 be the n^{th} term

\therefore 128 = 2 ( \sqrt{2} )^{n-1}

\Rightarrow 64 = ( \sqrt{2})^{n-1}

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

\Rightarrow n - 1 = 12

\Rightarrow n = 13

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

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

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

We know a_n = ar^{n-1}

Let 729 be the n^{th} term

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

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

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

\Rightarrow n - 1 = 11

\Rightarrow n = 12

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

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

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

We know a_n = ar^{n-1}

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

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

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

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

\Rightarrow n - 1 + 1 = 9

\Rightarrow n = 9

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

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

Answer:

Given series 18, -12, 8, \ldots

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

We know a_n = ar^{n-1}

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

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

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

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

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

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

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

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Question 8: Find the 4^{th} 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:

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

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

We know a_n = ar^{n-1}

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

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

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

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

Answer:

Given a_4 = 27 \Rightarrow ar^3 = 27       … … … … … i)

Given a_7 = 729 \Rightarrow ar^6 = 729       … … … … … ii)

Dividing ii) by i) we get

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

\Rightarrow r^3 = 27 

\Rightarrow r^3 = 3^3 

\Rightarrow r = 3 

Substituting in i) we get

a(3)^3 = 27 \Rightarrow a = 1 

Therefore the G.P. is 1, 3, 9, 27, \ldots 

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

Answer:

Given a_7 = a_4

\Rightarrow ar^6 = 8 ar^3

\Rightarrow r^3 = 8

\Rightarrow r = 2

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

Therefore the G.P. is 3, 6, 12, 24, \ldots

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

Answer:

For G.P. 5, 10, 20, \ldots 

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

We know a_n = ar^{n-1}

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

For G.P. 1280, 640, 320, \ldots

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

We know a_n = ar^{n-1}

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

From i) and ii) we get

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

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

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

\Rightarrow 2n - 2 = 8

\Rightarrow n = 5

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

Answer:

Given a_5 = ar^4 = p      … … … … … i)

a_8 = ar^7 = q      … … … … … ii)

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

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

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

Answer:

Given a_4 = (a_2)^2    and a = -3

\Rightarrow ar^3 = ( ar)^2

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

\Rightarrow r = a

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

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

Answer:

ar^2 = 24      … … … … … i)

ar^5 = 192      … … … … … ii)

Dividing ii) and i)  we get

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

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

Substituting in i) we get

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

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

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Question 15: If a, b, c and p are different real numbers such that

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

Answer:

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

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

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

\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,

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

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

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

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

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

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

Answer:

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

Take the first two terms

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

Apply componendo and dividendo we get

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

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

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

Similarly,  Take the last two terms

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

Apply componendo and dividendo we get

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

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

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

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

Therefore a, b, c and d are in G.P.

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

Answer:

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

a_q = ar^{q-1} = p      … … … … … ii)

Dividing i) by ii)

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

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

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

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

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

Substituting the value of r in ii)  we get

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

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

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

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

Substituting i) and ii) in iii) we get

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}

= \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]

= \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]

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

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

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

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