Geometric Progressions

EXERCISE 20.1

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,-1 / 2, \dots

Sol :

We have :

a_{1}=4,a_{2}=-2, a_{3}=1, a_{4}=-\frac{1}{2}

\frac{a_{2}}{a_{1}}=\frac{-2}{4}=\frac{-1}{2} , \frac{a_{3}}{a_{2}}=\frac{\phantom{-}1}{-2} , \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}

Thus, a_{1}, a_{2},a_{3} and a_{4} are in G.P., where a=4 and r=\frac{-1}{2}

 

(ii) -2 / 3,-6,-54, \ldots

Sol :

We have :

a_{1}=\frac{-2}{3},a_{2}=-6, a_{3}=-54

\frac{a_{2}}{a_{1}}=\frac{-6}{\frac{-2}{3}}=9,\frac{a_{3}}{a_{2}}=\frac{-54}{-6}=9

\therefore~\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=9

Thus, a_{1},a_{2} and a_{3} are in G.P., where a=\frac{-2}{3} and r=9

 

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

Sol :

We have :

a_{1}=a,a_{2}=\frac{3 a^{2}}{4},a_{3}=\frac{9 a^{3}}{16}

\frac{a_{2}}{a_{1}}=\frac{\frac{3 a^{2}}{4}}{a}=\frac{3 a}{4},\frac{a_{3}}{a_{2}}=\frac{\frac{9 a^{3}}{16}}{\frac{3 a^{2}}{4}}=\frac{3 a}{4}

\therefore~\frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{3 a}{4}

Thus, a_{1},a_{2} and a_{3} are in G.P., where the first term is a and the common ratio is \frac{3 a}{4} .

 

(iv)1 / 2,1 / 3,2 / 9,4 / 27, \ldots

Sol :

We have :

a_{1}=\frac{1}{2},a_{2}=\frac{1}{3},a_{3}=\frac{2}{9},a_{4}=\frac{4}{27}

\frac{a_{2}}{a_{1}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}, \frac{a_{3}}{a_{2}}=\frac{\frac{2}{9}}{\frac{1}{3}}=\frac{2}{3}, \frac{a_{4}}{a_{3}}=\frac{\frac{4}{27}}{\frac{2}{9}}=\frac{2}{3}

\therefore~ \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\frac{a_{4}}{a_{3}}=\frac{2}{3}

Thus, a_{1},a_{2}, a_{3} and a_{4} are in G.P., where the first term is \frac{1}{2} and the common ratio is \frac{2}{3} .

 

QUESTION 2

Show that the sequence <a_{n}> , defined by a_{n}=\frac{2}{3^{n}}, n \in N is a G.P.

Sol :

We have :

a_{n}=\frac{2}{3^{n}}, n \in N

Putting n = 1 , 2 , 3 , …..

a_{1}=\frac{2}{3^{1}}=\frac{2}{3}, a_{2}=\frac{2}{3^{2}}=\frac{2}{9},a_{3}=\frac{2}{3^{3}}=\frac{2}{27} and so on.

\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{3}}=\frac{1}{3}, \frac{a_{3}}{a_{2}}=\frac{\frac{2}{27}}{\frac{2}{9}}=\frac{1}{3} and so on.

\therefore~ \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\ldots=\frac{1}{3}

So, the sequence is an G.P., where \frac{2}{3} is the first term and \frac{1}{3} is the common ratio.

 

QUESTION 3

Find :

(i) the ninth term of the G.P. 1,4,16,64, \ldots

sol :

Here , 

First term, a=1

Common ratio, r=\frac{a_{2}}{a_{1}}=\frac{4}{1}=4

9 th term =a_{9}=a r^{(9-1)}

=1(4)^{8}=4^{8}

= 65536

 

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

sol :

Here ,

First term, a=\frac{-3}{\phantom(-)4}

Common ratio, r=\frac{a_{2}}{a_{1}}=\frac{\frac{1}{2}}{-\frac{3}{4}}=-\frac{2}{3}

10 th term =a_{10}=a r^{(10-1)} =\left(\frac{-3}{4}\right)\left(\frac{-2}{3}\right)^{9} =\frac{1}{2}\left(\frac{2}{3}\right)^{8}

Thus, the 10 th term of the given GP is \frac{1}{2}\left(\frac{2}{3}\right)^{8}

 

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

sol :

Here ,

First term , a = 0.3

Common ratio, r=\frac{a_{2}}{a_{1}}=\frac{0.06}{0.3}=0.2

8 th term =a_{8}=a r^{(8-1)} =0.3(0.2)^{7}

Thus, the 8 th term of the given GP is 0.3(0.2)^{7}

 

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

sol :

Here,

First term, a=\frac{1}{a^{3} x^{3}}

Common ratio, r=\frac{a_{2}}{a_{1}} =\frac{a x}{\frac{1}{a^{3} x^{3}}}=a^{4} x^{4}

12 th term =a_{12}=a r^{(12-1)} =\frac{1}{a^{3} x^{3}}\left(a^{4} x^{4}\right)^{11}=a^{41} x^{41}

Thus, the 12 th term of the given GP is a^{41} x^{41}

 

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

sol :

Here ,

First term, a=\sqrt{3}

Common ratio, r=\frac{a_{2}}{a_{1}} =\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}

n th term =a_{n}=a r^{(n-1)}=\sqrt{3}\left(\frac{1}{3}\right)^{n-1} Thus, the <em>n</em> th term of the given GP is\sqrt{3}\left(\frac{1}{3}\right)^{n-1}

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(vi) the 10 th term of the G.P.

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\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \ldots sol : Here , First term,a=\sqrt{2} Common ratio,r=\frac{a_{2}}{a_{1}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2} 10 th term=a_{10}=a r^{(10-1)}=\sqrt{2}\left(\frac{1}{2}\right)^{9}=\frac{1}{\sqrt{2}} \times \frac{1}{2^{8}} Thus, the 10 th term of the given GP is\frac{1}{\sqrt{2}} \times \frac{1}{2^{8}}

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<span style="background-color: #ffff00;"><strong>QUESTION 4</strong></span>
Find the 4 th term from the end of the G.P.

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\frac{2}{27}, \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162 sol : first term,a=\frac{2}{27} Common ratio,r=\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{27}}=3 Last term, <em>l = 162</em> After reversing the given G.P., we get another G.P. whose first term island common ratio is\frac{1}{r}  4 th term from the end=l\left(\frac{1}{r}\right)^{4-1}=(162)\left(\frac{1}{3}\right)^{3}=6

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ALTERNATE METHOD
 
First term ,

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a=\dfrac{2}{27} Common ratio,r=\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{27}}=3

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last term , l = 162
To get the the position from front :
( total terms - position from end + 1 )
 
Total terms :


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a_n=ar^{n-1} 162=\dfrac{2}{27}\times{3}^{n-1} \dfrac{162\times 27}{2}=3^{n-1} 81\times 27=3^{n-1} 3^7=3^{n-1} Comparing the power of both the sides n-1=7 n = 8 ( total terms - position from end + 1 )  = 8 - 4 + 1 = 5 Which means 5 position from starting a_{n}=ar^{n-1} a_{5}=\dfrac{2}{27}\times 3^{5-1} a_{5}=\dfrac{2}{3^3}\times {3^4} a_{5}=2\times 3

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= 6
 
<span style="background-color: #ffff00;"><strong>QUESTION 5</strong></span>
Which term of the progression

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0.004,0.02,0.1, \ldotsis 12.5 ? Sol: We have : \frac{a_{2}}{a_{1}}=\frac{0.02}{0.004}=5, \frac{a_{3}}{a_{2}}=\frac{0.1}{0.02}=5 Common ratio\Rightarrow \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=5 The given progression is a G.P. whose first term, <em>a</em> is 0.004 and common ratio, <em>r</em> is 5 Let the <em>n</em> th term be 12.5 \therefore a_{n}=12.5 \Rightarrow ~a r^{n-1}=a_n \Rightarrow ~(0.004)\times 5^{n-1}=12.5 \Rightarrow(5)^{n-1}=\frac{12.5}{0.004} \Rightarrow(5)^{n-1}=3125 \Rightarrow(5)^{n-1}=(5)^{5} Comparing the power of both the sides \Rightarrow n-1=5

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<em>n = 6</em>
Thus, 6 th term of the given G.P. is 12.5
 
<span style="background-color: #ffff00;"><strong>QUESTION 6</strong></span>
Which term of the G.P.:
(i)

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\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{4 \sqrt{2}}, \ldotsis\frac{1}{512 \sqrt{2}} ? Sol : Here, first term,a=\sqrt{2} Common ratior=\frac{1}{2} Let then^{th}term be\frac{1}{512 \sqrt{2}} \therefore~a_{n}=\frac{1}{512 \sqrt{2}} \Rightarrow a r^{n-1}=a_{n} \Rightarrow \sqrt{2}\times \dfrac{1}{2}^{n-1}=\frac{1}{512 \sqrt{2}} \Rightarrow\left(\frac{1}{2}\right)^{n-1}=\frac{1}{1024} \Rightarrow\left(\frac{1}{2}\right)^{n-1}=\left(\frac{1}{2}\right)^{10} \Rightarrow n-1=10 <em>n = 11</em> Thus, the 11^{\text { th }} term of the given G.P. is \frac{1}{512 \sqrt{2}}

 

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

Sol :

first term, a=2

common ratio, r=\sqrt{2}

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

\therefore a_{n}=128

\Rightarrow a r^{n-1}=a_n

\Rightarrow 2 \times \sqrt{2}^{n-1}=128

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

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

\Rightarrow n-1=12

n = 13

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

 

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

Sol :

first term, a=\sqrt{3}

common ratio, r=\sqrt{3}

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

\therefore a_{n}=729

\Rightarrow a r^{n-1}=a_{n}

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

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

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

\Rightarrow n-1=11

n = 12

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

 

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

Sol :

first term, a=\frac{1}{3}

common ratio, r=\frac{1}{3}

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

\therefore a_{n}=\frac{1}{19683}

\Rightarrow \quad a r^{n-1}=a_{n}

\Rightarrow \quad\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)^{n-1}=\frac{1}{19683}

\Rightarrow\left(\frac{1}{3}\right)^{n-1}=\frac{3}{(3)^{9}}

\Rightarrow\left(\frac{1}{3}\right)^{n-1}=\left(\frac{1}{3}\right)^{8}

\Rightarrow n-1=8

n = 9

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

 

QUESTION 7

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

Sol :

first term, a = 18

common ratio, r=\frac{-2}{\phantom{-}3}

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

a r^{n-1}=a_{n}

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

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

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

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

\Rightarrow n-1=8

n = 9

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

 

QUESTION 8

Find the 4 th term from the end of the G.P. \frac{1}{2}, \frac{1}{2}, \frac{1}{18}, \frac{1}{54}, \ldots, \frac{1}{4374}

Sol :

After reversing the given G.P., we get another G.P.

whose first term, l is \frac{1}{4374}

common ratio is 3

4^{\text { th }} term from the end =l\left(\frac{1}{r}\right)^{4-1}

=\left(\frac{1}{4374}\right)(3)^{4-1}

=\left(\frac{27}{4374}\right)

=\frac{1}{162}

 

QUESTION 9

The fourth term of a G.P. is 27 and the 7 th term is 729, find the G.P.

Sol :

Let a be the first term and r be the common ratio of the given G.P.

a_{4}=27 and a_{7}=729

\Rightarrow a r^{3}=27 and a r^{6}=729

\Rightarrow \frac{a r^{6}}{a r^{3}}=\frac{729}{27}

\Rightarrow r^{3}=3^{3}

r = 3

Putting r=3 in a r^{3}=27

a(3)^{3}=27

a = 1

Thus, the given G.P. is 1,3,9, \dots

 

QUESTION 10

The seventh term of a G.P. is 8 times the fourth term and 5th term is 48 . Find the G.P.

Sol :

Let a be the first term and r be the common ratio.

a_7=4\times a_4 and a_{5}=48

\Rightarrow a r^{6}=8 a r^{3} and a r^{4}=48

\Rightarrow r^{3}=8

\Rightarrow r^{3}=2^{3}

r = 2

Putting r=2 in a r^{4}=48

\Rightarrow~(a)(2)^{4}=48

a = 3

Thus, the given G.P. is 3,6,12, \dots

 

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

Sol :

First term, a=5

Common ratio, r=2

a_{n}=(a)(r)^{n-1}

a_{n}=(5)(2)^{n-1}\dots (i)

Similarly, a_{n}=(1280)\left(\frac{1}{2}\right)^{n-1}\dots(ii)

From (i) and (ii)

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

\Rightarrow \frac{1}{256}=\left(\frac{1}{4}\right)^{n-1}

\Rightarrow\left(\frac{1}{4}\right)^{4}=\left(\frac{1}{4}\right)^{n-1}

\Rightarrow n-1=4

n = 5

 

QUESTION 12

If 5^{\text { th }}, 8^{\text { th }} and 11^{\text { th }} terms of a G.P. are p , q and s respectively, prove that q^{2}=p s

Sol :

Let a be the first term and r be the common ratio of the given G.P.

p=5^{\mathrm{th}} term

\Rightarrow p=a r^{4} \quad \ldots(1)

q=8^{\text { th }} term

\Rightarrow q=a r^{7} \quad \ldots(2)

s=11^{\text { th }}

\Rightarrow s=a r^{10} \quad \ldots \quad(3)

Now , q^{2}=\left(a r^{7}\right)^{2}=a^{2} r^{14}

\Rightarrow\left(a r^{4}\right)\left(a r^{10}\right)=p s [from(1) and (2)]

\therefore \quad q^{2}=p s

 

QUESTION 13

The 4 th term of a G.P. is square of its second term, and the first term is -3  . Find its 7^{\text { th }} term.

Sol :

Let r be the common ratio of the given G.P.

Then, a_{4}=\left(a_{2}\right)^{2} [Given]

Now , a r^{3}=a^{2} r^{2}

r = a

\Rightarrow r=-3 [ Putting a = -3]

\therefore a_{7}=a r^{6}

\Rightarrow a_{7}=(-3)(-3)^{6} [ Putting a = -3 and  r = -3]

\Rightarrow a_{7}=(-3)(-729)

\Rightarrow a_{7}=-2187

Thus, the 7^{\text { th }} term of the G.P. is -2187

 

QUESTION 14

In a GP the 3^{\text { rd }} term is 24 and the 6^{\text { th }} term is 192 . Find the 10^{\text { th }} term.

Sol :

Let a be the first term and r be the common ratio.

\therefore a_{3}=24 and a_{6}=192

\Rightarrow a r^{2}=24 and a r^{5}=192

\Rightarrow \frac{a r^{5}}{a r^{2}}=\frac{192}{24}

\Rightarrow r^{3}=8

\Rightarrow r^{3}=2^{3}

r = 2

Putting r=2 in a r^{2}=24

a(2)^{2}=24

a = 6

Now, 10^{\text { th }} term =a_{10}=a r^{9}

Putting a=6 and r=2 in a_{10}=a r^{9}

\Rightarrow a_{10}=(6)(2)^{9}

= 3072

Thus, the 10^{\text { th }} term of the G.P. is 3072

 

QUESTION 15

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

\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \leq 0, then show that a, b, c and d are in G.P.

Sol :

\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+d^{2}\right) \leq 0

\Rightarrow\left(a^{2} p^{2}+b^{2} p^{2}+c^{2} p^{2}\right)-2(a b p+b c p+c d p)+\left(b^{2}+c^{2}+d^{2}\right) \leq 0

\Rightarrow\left(a^{2} p^{2}-2 a b p+b^{2}\right)+\left(b^{2} p^{2}-2 b c p+c^{2}\right)+\left(c^{2} p^{2}-2 c d p+d^{2}\right) \leq 0

\Rightarrow(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2} \leq 0

\Rightarrow(a p-b)^{2}+(b p-c)^{2}+(c p-d)^{2}=0

\Rightarrow(a p-b)^{2}=0

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

Also, (b p-c)^{2}=0

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

Similiarly, \Rightarrow(c p-d)^{2}=0

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

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

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

 

QUESTION 16

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

Sol :

Given :

\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}

Now ,\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}

Applying componendo and dividendo

\Rightarrow \frac{(a+b x)+(a-b x)}{(a+b x)-(a-b x)}=\frac{(b+c x)+(b-c x)}{(b+c x)-(b-c x)}

\Rightarrow \frac{2 a}{2 b x}=\frac{2 b}{2 c x}

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

Similiarly, \frac{(b+c x)+(b-c x)}{(b+c x)-(b-c x)}=\frac{(c+d x)+(c-d x)}{(c+d x)-(c-d x)}

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

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

 

QUESTION 17

If the pth and qth terms of a G.P. are q and p , respectively, then show that (p+q) th term is \left(\frac{q^{p}}{p^{q}}\right)^{\frac{1}{p-q}}

Sol :

As , a_{p}=q

\Rightarrow a r^{(p-1)}=q \quad \ldots \ldots(\mathrm{i})

Also, a_{q}=p

\Rightarrow a r^{(q-1)}=p \quad \ldots \ldots(\mathrm{ii})

Dividing (i) by (ii) , we get

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

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

\Rightarrow r^{(p-q)}=\frac{q}{p}

\Rightarrow r=\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}

Substituting the value of r in (ii), we get

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

\Rightarrow a\left[\left(\frac{q}{p}\right)^{\frac{(q-1)}{(p-q)}}\right]=p

\Rightarrow a=p \times\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}}

\Rightarrow a=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}}

Now ,

a_{(p+q)}=a r^{(p+q-1)}

=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}} \times\left[\left(\frac{q}{p}\right)^{\frac{1}{(p-q)}}\right]^{(p+q-1)}

=p\left(\frac{p}{q}\right)^{\frac{(q-1)}{(p-q)}} \times\left(\frac{q}{p}\right)^{\frac{(p+q-1)}{(p-q)}}

=p\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}} \times\left(\frac{q}{p}\right)^{\frac{(p+q-1)}{(p-q)}}

=p \times\left(\frac{q}{p}\right)^{\frac{-(q-1)}{(p-q)}+\frac{(p+q-1)}{(p-q)}}

=p \times\left(\frac{q}{p}\right)^{\frac{-q+1+p+q-1}{(p-q)}}

=p \times\left(\frac{q}{p}\right)^{\frac{p}{(p-q)}}

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

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

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

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

=\frac{q \times \frac{1}{(p-q)}}{p \times \frac{1}{(p-q)}}

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

 

 

 

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