# EXERCISE 20.3

**QUESTION 1**

Find the sum of the following geometric progressions:

(i) to 7 terms;

Sol :

Here, *a=2* and *r=3*

= 2787 – 1

= 2186

(ii) to 8 terms

Sol :

Here, *a=1* and *r=3*

= 3280

(iii) to 9 terms;

Sol :

Here, *a=1* and

(iv) to *n* terms

Sol :

Here, and

(v) to 10 terms.

Sol :

Here, *a=4* and

**QUESTION 2**

Find the sum of the following geometric series:

(i) to 8 terms;

Sol :

Here, *a=0.15* and

(ii) to 8 terms

Sol :

Here, and

(iii) to 5 terms;

Sol :

Here, and

(iv) to *n* termsย

Sol :

(v)ย to 2 terms

Sol :

(vi)

Sol :

(vii) to *n* terms

Sol :

(viii) to *n* terms

Sol :

(ix) to *n* terms

Sol :

**QUESTION 3**

Evaluate the following :

(i)

Sol :

= 22 + 265719

= 265741

(ii)

Sol :

(iii)

Sol :

**QUESTION 4**

Find the sum of the following series:

(i) to terms;

Sol :

taking 5 as common :

terms

terms

times

(ii) to terms

Sol :

Taking 7 as common :

terms

n terms

times

(iii) to terms

Sol :

This can be rewritten as :

times

(iv) to terms.

Sol :

Taking 5 as common :

terms

to *n* terms )

=\dfrac{5}{9}\left\{n-\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\dfrac{1}{10^3}+\dots{\text{ n terms }}\right)\right}

(v) to terms

Sol :

Taking 6 as common :

=\dfrac{6}{9}\left\{n-\left(\dfrac{1}{10}+\dfrac{1}{10^2}+\dfrac{1}{10^3}+\dots{\text{ n terms }}\right)\right}

**QUESTION 5**

How many terms of the G.P. be taken together to make ?

Sol :

First term , *a = 3*

Common ratio,

*n = 10*

**QUESTION 6**

How many terms of the series must be taken to make the sum equal to 728 ?

Sol :

First term , *a = 2*

Common ratio, *r = 3*

Given :

n = 6

**QUESTION 7**

How many terms of the sequence must be taken to make the sum ?

Sol :

First term ,

Common ratio,

*n = 6*

**QUESTION 8**

The sum of *n* terms of the G.P. is 381 . Find the value of *n*

Sol :

First term , *a = 3*

Common ratio, *r = 3*

Sum of *n* terms,

*n = 7*

**QUESTION 9**

The common ratio of a G.P. is 3 and the last term is 486 . If the sum of these terms be 728, find the first term.

Sol :

common ratio, *r = 3*

term,

(i)

Now,

[from(i)]

*a = 1458 – 1456*

*a = 2*

**QUESTION 10**

The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152 . Find the common ratio.

Sol :

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

and

Then, according to the question

Now, let

Now, applying the quadratic formula

or

or

or

But, is not possible.

**QUESTION 11**

The 4 th and 7 th terms of a G.P. are and respectively. Find the sum of *n* terms of the G.P.

Sol :

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

**QUESTION 12**

**QUESTION 13**

**QUESTION 14**

**QUESTION 15**

**QUESTION 16**