# 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)

Sol :

We have :

,, ,

, ,

Thus, , , and are in G.P., where a=4 and

(ii)

Sol :

We have :

,,

,

Thus, , and are in G.P., where and

(iii)

Sol :

We have :

,,

,

Thus, , and are in G.P., where the first term is a and the common ratio is .

(iv)

Sol :

We have :

,,,

, ,

Thus, , and are in G.P., where the first term is and the common ratio is .

QUESTION 2

Show that the sequence , defined by is a G.P.

Sol :

We have :

Putting n = 1 , 2 , 3 , …..

, , and so on.

, and so on.

So, the sequence is an G.P., where is the first term and is the common ratio.

QUESTION 3

Find :

(i) the ninth term of the G.P.

sol :

Here ,

First term, a=1

Common ratio,

9 th term

= 65536

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

sol :

Here ,

First term,

Common ratio,

10 th term

Thus, the 10 th term of the given GP is

(iii) the 8 th term of the G.P.

sol :

Here ,

First term , a = 0.3

Common ratio,

8 th term

Thus, the 8 th term of the given GP is 0.3

(iv) the 12 th term of the G.P.

sol :

Here,

First term,

Common ratio,

12 th term

Thus, the 12 th term of the given GP is

(v) n th term of the G.P.

sol :

Here ,

First term,

Common ratio,

n th term =a_{n}=a r^{(n-1)}=\sqrt{3}\left(\frac{1}{3}\right)^{n-1}\sqrt{3}\left(\frac{1}{3}\right)^{n-1}

*** QuickLaTeX cannot compile formula:

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

*** Error message:
Missing $inserted.  \sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \ldotsa=\sqrt{2}r=\frac{a_{2}}{a_{1}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}=a_{10}=a r^{(10-1)}=\sqrt{2}\left(\frac{1}{2}\right)^{9}=\frac{1}{\sqrt{2}} \times \frac{1}{2^{8}}\frac{1}{\sqrt{2}} \times \frac{1}{2^{8}} *** QuickLaTeX cannot compile formula: <span style="background-color: #ffff00;"><strong>QUESTION 4</strong></span> Find the 4 th term from the end of the G.P. *** Error message: Missing$ inserted.



\frac{2}{27}, \frac{2}{27}, \frac{2}{9}, \frac{2}{3}, \ldots, 162a=\frac{2}{27}r=\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{27}}=3l\frac{1}{r}=l\left(\frac{1}{r}\right)^{4-1}=(162)\left(\frac{1}{3}\right)^{3}=6

*** QuickLaTeX cannot compile formula:

ALTERNATE METHOD

First term ,

*** Error message:
Missing $inserted.  a=\dfrac{2}{27}r=\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{2}{27}}=3 *** QuickLaTeX cannot compile formula: last term , l = 162 To get the the position from front : ( total terms - position from end + 1 ) Total terms : *** Error message: Missing$ inserted.



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}n-1=7a_{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

*** QuickLaTeX cannot compile formula:

= 6

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

*** Error message:
Missing $inserted.  0.004,0.02,0.1, \ldots\frac{a_{2}}{a_{1}}=\frac{0.02}{0.004}=5 \frac{a_{3}}{a_{2}}=\frac{0.1}{0.02}=5\Rightarrow \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=5\therefore a_{n}=12.5\Rightarrow ~a r^{n-1}=a_n\Rightarrow ~(0.004)\times 5^{n-1}=12.5\Rightarrow(5)^{n-1}=\frac{12.5}{0.004}\Rightarrow(5)^{n-1}=3125\Rightarrow(5)^{n-1}=(5)^{5}\Rightarrow n-1=5 *** QuickLaTeX cannot compile formula: <em>n = 6</em> Thus, 6 th term of the given G.P. is 12.5 <span style="background-color: #ffff00;"><strong>QUESTION 6</strong></span> Which term of the G.P.: (i) *** Error message: Missing$ inserted.



\sqrt{2}, \frac{1}{\sqrt{2}}, \frac{1}{2 \sqrt{2}}, \frac{1}{4 \sqrt{2}}, \ldots\frac{1}{512 \sqrt{2}} ?a=\sqrt{2}r=\frac{1}{2}n^{th}\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 term of the given G.P. is

(ii) is 128 ?

Sol :

first term, a=2

common ratio,

Let the term be 128

n = 13

Thus, the term of the given G.P. is 128

(iii) is 729 ?

Sol :

first term,

common ratio,

Let the term be 729

n = 12

Thus, the term of the given G.P. is 729

(iv) is ?

Sol :

first term,

common ratio,

Let the term be

n = 9

Thus, the term of the given G.P. is

QUESTION 7

Which term of the progression is ?

Sol :

first term, a = 18

common ratio,

Let the term be

n = 9

Thus, the term of the given G.P. is

QUESTION 8

Find the 4 th term from the end of the G.P.

Sol :

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

whose first term, l is

common ratio is 3

term from the end

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.

and

and

r = 3

Putting in

a = 1

Thus, the given G.P. is

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 be the first term and be the common ratio.

and

and

r = 2

Putting r=2 in

a = 3

Thus, the given G.P. is

QUESTION 11

If the G.P’s and have their n th terms equal, find the value of n

Sol :

First term,

Common ratio,

Similarly,

From (i) and (ii)

n = 5

QUESTION 12

If and terms of a G.P. are p , q and s respectively, prove that

Sol :

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

term

term

Now

[from(1) and (2)]

QUESTION 13

The 4 th term of a G.P. is square of its second term, and the first term is -3  . Find its term.

Sol :

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

Then, [Given]

Now

r = a

[ Putting a = -3]

[ Putting a = -3 and  r = -3]

Thus, the term of the G.P. is

QUESTION 14

In a GP the term is 24 and the term is 192 . Find the term.

Sol :

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

and

and

r = 2

Putting in

a = 6

Now, term

Putting a=6 and r=2 in

= 3072

Thus, the term of the G.P. is 3072

QUESTION 15

If and are different real numbers such that:

then show that a, b, c and d are in G.P.

Sol :

Also,

Similiarly,

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

QUESTION 16

If then show that a, b, c and d are in G.P.

Sol :

Given :

Now ,

Applying componendo and dividendo

Similiarly,

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 th term is

Sol :

As

Also,

Dividing (i) by (ii) , we get

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

Now ,