Ch. 1_Arithmetic Progression And

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SCHOOL SECTION 1

o Introduction ::

We have seen different types of numbers since our childhood. Like set ofpositive even numbers, odd numbers etc. In all these sets some specificpattern is followed. In nature also we see that some pattern is followed e.g.

Arrangement of petals in flower or arrangement in fruits or in tree.

All these arrangements in nature look beautiful because some pattern isfollowed in all of them.

o Sequence :

A sequence is a collection of numbers arranged in a definite order accordingto some definite rule.Each number in the sequence is called a term of the sequence.

The number in the first position is called the first term and it is denoted by t1.

Similarly the number in the second and third positions are denoted as t2and t3

respectively.

In general, the number in the nth position is called nth term and is denotedby t

n. A sequence is usually denoted by { t

n} or < t

n> and read as sequence t

n.

o Examples of sequences :

1. 2, 4, 6, 8, ...... Here t 1

= 2, t2

= 4, t3

= 6, t4

= 8, .......2. 3, 6, 9, 12, ...... Here t

1= 3, t

2= 6, t

3= 9, t

4= 12, .......

3. 5, 25, 125, 625, ...... Here t 1

= 5, t2

= 25, t3

= 125, t4

= 625, .......4. 4, 2, 0, 2, ...... Here t

1= 4, t

2= 2, t

3= 0, t

4= 2, .......

5.1

2

,1

6

,1

18

,1

54

, ...... Here t 1

=1

2

, t2

=1

6

, t3

=1

18

, t4

=1

54

, .......

o Sum of first n terms of a sequence :S

1= t

1

S2

= t1

+ t2

S3

= t1

+ t2

+ t3

S4

= t1

+ t2

+ t3

+ t4

In general Sn

= t1

+ t2

+ t3

+ ....... + tn.

From the above equations we observed thatS

1= t

1

S2

S1

= t2

S3

S2

= t3

In general Sn Sn1 = tn. t

n= S

n S

n1.

Arithmetic Progression AndGeometric Progression

1.


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SCHOOL SECTION2

o Types of sequence :

If the number of terms in a sequence is finite then it is called a finitesequence. e.g. : 5, 9, 13, 17.

If the number of terms in the sequence is not finite then it is called

infinite sequence. e.g. : 7, 10, 13, 16, ......

EXERCISE - 1.1 (TEXT BOOK PAGE NO. 4) :

1. For each sequence, find the next four terms.(i) 1, 2, 4, 7, 11, ..... (1 mark)Sol. t

1= 1 + 0 = 1

t2

= 1 + 1 = 2t

3= 2 + 2 = 4

t4

= 4 + 3 = 7t

5= 7 + 4 = 11

t6

= 11 + 5 = 16t

7

= 16 + 6 = 22t

8= 22 + 7 = 29

t9

= 29 + 8 = 37

The next four terms of the sequence are 16, 22, 29 and 37.

(ii) 3, 9, 27, 81, ..... (1 mark)Sol. t

1= 31 = 3

t2

= 32 = 9t

3= 33 = 27

t4

= 34 = 81t

5= 35 = 243

t6

= 36 = 729

t7 = 37 = 2187t

8= 38 = 6561


(iii) 1, 3, 7, 15, 31, .... (1 mark)Sol. t

1= 0 + 20 = 0 + 1 = 1

t2

= 1 + 21 = 1 + 2 = 3t

3= 3 + 22 = 3 + 4 = 7

t4

= 7 + 23 = 7 + 8 = 15t

5= 15 + 24 = 15 + 16 = 31

t6

= 31 + 25 = 31 + 32 = 63

t7 = 63 + 26

= 63 + 64 = 127t

8= 127 + 27 = 127 + 128 = 255

t9

= 255 + 28 = 255 + 256 = 511


(iv) 192, 96, 48, 24, .... (1 mark)Sol. t

1= 192

t2

=192

2 = 96

t3

=96

2 = 48

t4

=48

2 = 24


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t5

=24

2 = 12

t6

=12

2 = 6

t7

=6 2 = 3

t8

=3

2 =3

2

The next four terms of the sequence are 12, 6, 3 and3

2

.

(v) 2, 6, 12, 20, 30, ... (1 mark)Sol. t

1= 0 + 2 = 2

t2

= 2 + 4 = 6

t3 = 6 + 6 = 12t

4= 12 + 8 = 20

t5

= 20 + 10 = 30t

6= 30 + 12 = 42

t7

= 42 + 14 = 56t

8= 56 + 16 = 72

t9

= 72 + 18 = 90


(vi) 0.1, 0.01, 0.001, 0.0001, .... (1 mark)Sol. t

1= 0.1

t2

= 0.110

= 0.01

t3

=0.01

10= 0.001

t4

=0.001

10= 0.0001

t5

=0.0001

10= 0.00001

t6

=0.00001

10= 0.000001

t7

=0.000001

10= 0.0000001

t8

=0.0000001

10= 0.00000001

The next four terms of the sequence are 0.00001, 0.000001,0.0000001 and 0.00000001.

(vii) 2, 5, 8, 11, .... (1 mark)Sol. t

1= 2

t2

= 2 + 3 = 5

t3 = 5 + 3 = 8t

4= 8 + 3 = 11


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SCHOOL SECTION4

t5

= 11 + 3 = 14t

6= 14 + 3 = 17

t7

= 17 + 3 = 20t

8= 20 + 3 = 23


(viii) 25, 23, 21, 19, ..... (1 mark)Sol. t

1= 25

t2

= 25 + 2= 23t

3= 23 + 2= 21

t4

= 21 + 2= 19t

5= 19 + 2= 17

t6

= 17 + 2= 15t

7= 15 + 2= 13

t8

= 13 + 2= 11


(ix) 2, 4, 8, 16, ..... (1 mark)Sol. t

1= 21 = 2

t2

= 22 = 4t

3= 23 = 8

t4

= 24 = 16t

5= 25 = 32

t6

= 26 = 64t

7= 27 = 128

t8

= 28 = 256


(x)1

2,

1

6,

1

18,

1

54, .... (2 mark)

Sol. t1

=1

2

t2

=1 1

2 3 =

1

6

t3

=1 1

6 3 =

1

18

t4 =

1 1

18 3

=

1

54

t5

=1 1

54 3 =

1

162

t6

=1 1

162 3 =

1

486

t7

=1 1

486 3 =

1

1458

t8

=1 1

1458 3 =

1

4374

The next four terms of the sequence are 1162

, 1486

, 11458

and 14374

.


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SCHOOL SECTION 5


2. Find the first five terms of the following sequences, whose nth termsare given :

(i) tn

= 4n 3 (1 mark)Sol. t

n= 4n 3

t1 = 4 (1) 3 = 4 3 = 1 t

2= 4 (2) 3 = 8 3 = 5

t3

= 4 (3) 3 = 12 3 = 9 t

4= 4 (4) 3 = 16 3 = 13

t5

= 4 (5) 3 = 20 3 = 17

The first five terms of the sequence are 1, 5, 9, 13 and 17.

(ii) tn

= 2n 5 (1 mark)Sol. t

n= 2n 5

t1

= 2 (1) 5 = 2 5 = 3 t

2= 2 (2) 5 = 4 5 = 1

t3 = 2 (3) 5 = 6 5 = 1 t4

= 2 (4) 5 = 8 5 = 3 t

5= 2 (5) 5 = 10 5 = 5


(iii) tn

= n + 2 (1 mark)Sol. t

n= n + 2

t1

= 1 + 2 = 3 t

2= 2 + 2 = 4

t3

= 3 + 2 = 5 t

4= 4 + 2 = 6

t5

= 5 + 2 = 7


(iv) tn

= n2 2n (1 mark)Sol. t

n= n2 2n

t1

= 1 2 (1) = 1 2 = 1 t

2= 22 2 (2) = 4 4 = 0

t3

= 32 2 (3) = 9 6 = 3 t

4= 42 2 (4) = 16 8 = 8

t5

= 52 2 (5) = 25 10 = 15


(v) tn = n3

(1 mark)Sol. tn

= n3

t1

= 13 = 1 t

2= 23 = 8

t3

= 33 = 27 t

4= 43 = 64

t5

= 53 = 125


(vi) tn

=+

1

n 1(1 mark)

Sol. tn

= 1n 1+


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SCHOOL SECTION6

t1

=1

1 1+ =1

2

t2

=1

2 1+ =1

3

t3 =

1

3 1+ =

1

4

t4

=1

4 1+ =1

5

t5

=1

5 1+ =1

6

The first five terms of the sequence are1

2,

1

3,

1

4,

1

5and

1

6.


3. Find the first three terms of the sequences for which Snis

given below :

(i) Sn = n2

(n + 1) (2 marks)Sol. S

n= n2 (n + 1)

S1

= 12 (1 + 1) = 1 (2) = 2 S

2= 22 (2 + 1) = 4 (3) = 12

S3

= 32 (3 + 1) = 9 (4) = 36We know that,t

1= S

1= 2

t2

= S2

S1

= 12 2 = 10t

3= S

3 S

2= 36 12 = 24

The first three terms of the sequence are 2, 10 and 24.

(ii) Sn = +2 2

n (n 1)4

(2 marks)

Sol. Sn

=n (n 1)

4

+2 2

S1

=1 (1 1)

4

+2 2

=1 (2)

4

2

=1 4

4

= 1

S2

=2 (2 1)

4

+2 2

=4 (3)

4

2

= 9

S3

=3 (3 1)

4

+2 2

=9 4

4

2

= 9 4 = 36

We know that,t1

= S1

= 1t

2= S

2 S

1= 9 1 = 8

t3

= S3

S2

= 36 9 = 27


(iii)+ +n (n 1) (2n 1)

6(2 marks)

Sol. Sn

=n (n 1) (2n 1)

6

+ +

S1

= 1 (1 1) [2 (1) 1]6

+ + = 1 2 36

= 66

= 1


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SCHOOL SECTION 7

S2

=2 (2 1) [2 (2) 1]

6

+ +=

2 3 5

6

=

30

6= 5

S3

=3 (3 1) [2 (3) 1]

6

+ +=

3 4 7

6

= 14

We know that,

t1 = S1 = 1t

2= S

2 S

1= 5 1 = 4

t3

= S3

S2

= 14 5 = 9


o Progression :Whenever we write some numbers one after the other, keeping a fix relationbetween two consecutive terms then we say the numbers are in progression.e.g. 1, 5, 10, 19, ..........

o Arithmetic Progression :

Arithmetic Progression is a sequence in which the difference between twoconsecutive terms is a constant.e.g. 4, 8, 12, 16, ...........

Here, the difference between any two consecutive terms is 4 which is aconstant. Therefore, the sequence is an A.P.e.g. 2, 4, 8, 16, ...........

Here, the difference between any two consecutive terms is not a constant.Therefore, the sequence is not an A.P.


1. Which of the following lists of numbers are Arithmetic Progressions?Justify.(i) 1, 3, 6, 10, ..... (1 mark)Sol. t

1= 1, t

2= 3, t

3= 6, t

4= 10

t2

t1

= 3 1 = 2t3

t2

= 6 3 = 3t4

t3

= 10 6 = 4 The difference between two consecutive terms is not constant. The sequence is not an A.P.

(ii) 3, 5, 7, 9, 11, ..... (1 mark)Sol. t

1= 3, t

2= 5, t

3= 7, t

4= 9, t

5= 11

t2 t1 = 5 3 = 2t3

t2

= 7 5 = 2t4

t3

= 9 7 = 2t5

t4

= 11 9 = 2 The difference between two consecutive terms is 2 which is constant. The sequence is an A.P.

(iii) 1, 4, 7, 10, .... (1 mark)Sol. t

1= 1, t

2= 4, t

3= 7, t

4= 10

t2

t1

= 4 1 = 3t3

t2

= 7 4 = 3t4

t3

= 10 7 = 3

The difference between two consecutive terms 3 which is constant. The sequence is an A.P.


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SCHOOL SECTION8

(iv) 3, 6, 12, 24, .... (1 mark)

Sol. t1

= 3, t2

= 6, t3

= 12, t4

= 24

t2

t1

= 6 3 = 3

t3

t2

= 12 6 = 6

t4

t3

= 24 12 = 12

The difference between two consecutive terms is not constant. The sequence is not an A.P.

(v) 22, 26, 28, 31, ... (1 mark)

Sol. t1

= 22, t2

= 26, t3

= 28, t4

= 31

t2

t1

= 26 22 = 4

t3

t2

= 28 26 = 2

t4

t3

= 21 28 = 3

The difference between two consecutive terms is not constant.

The sequence is not an A.P.

(vi) 0.5, 2, 3.5, 5, ... (1 mark)Sol. t

1= 0.5, t

2= 2, t

3= 3.5, t

4= 5

t2

t1

= 2 0.5 = 1.5

t3

t2

= 3.5 2 = 1.5

t4

t3

= 5 3.5 = 1.5

The difference between two consecutive terms is 1.5 which is constant.

The sequence is an A.P.

(vii) 4, 3, 2, 1, .... (1 mark)Sol. t

1= 4, t

2= 3, t

3= 2, t

4= 1,

t2

t1

= 3 4 = 1

t3 t2 = 2 3 = 1t4

t3

= 1 2 = 1

The difference between two consecutive terms is 1 which is constant.


(viii) 10, 13, 16, 19, ..... (1 mark)

Sol. t1

= 10, t2

= 13, t3

= 16, t4

= 19

t2

t1

= 13 ( 10) = 13 + 10 = 3

t3

t2

= 16 ( 13) = 16 + 13 = 3

t4

t3

= 19 ( 16) = 19 + 16 = 3

The difference between two consecutive terms is 3 which is constant.



2. Write the first five terms of the following Arithmetic Progressions where,

the common difference d and the first term a are given :

(i) a = 2, d = 2.5 (1 mark)

Sol. a = 2, d = 2.5

Here, t1

= a = 2

t2

= t1

+ d = 2 + 2.5 = 4.5

t3

= t2

+ d = 4.5 + 2.5 = 7

t4

= t3

+ d = 7 + 2.5 = 9.5

t5 = t4 + d = 9.5 + 2.5 = 12 The first five terms of the A.P. are 2, 4.5, 7, 9.5 and 12.


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SCHOOL SECTION 9

(ii) a = 10, d = 3 (1 mark)

Sol. a = 10, d = 3

Here, t 1

= a = 10

t2

= t1

+ d = 10 + ( 3) = 10 3 = 7

t3

= t2

+ d = 7 + ( 3) = 7 3 = 4

t4

= t3

+ d = 4 + ( 3) = 4 3 = 1

t5

= t4

+ d = 1 + ( 3) = 1 3 = 2

The first five terms of the A.P. are 10, 7, 4, 1 and 2.

(iii) a = 4, d = 0 (1 mark)

Sol. a = 4, d = 0

Here, t 1

= a = 4

t2

= t1

+ d = 4 + 0 = 4

t3

= t2

+ d = 4 + 0 = 4

t4

= t3

+ d = 4 + 0 = 4

t5 = t4 + d = 4 + 0 = 4 The first five terms of the A.P. are 4, 4, 4, 4 and 4.

(iv) a = 5, d = 2 (1 mark)

Sol. a = 5, d = 2

Here, t 1

= a = 5

t2

= t1

+ d = 5 + 2 = 7

t3

= t2

+ d = 7 + 2 = 9

t4

= t3

+ d = 9 + 2 = 11

t5

= t4

+ d = 11 + 2 = 13


(v) a = 3, d = 4 (1 mark)

Sol. a = 3, d = 4

Here, t 1

= a = 3

t2

= t1

+ d = 3 + 4 = 7

t3

= t2

+ d = 7 + 4 = 11

t4

= t3

+ d = 11 + 4 = 15

t5

= t4

+ d = 15 + 4 = 19


(vi) a = 6, d = 6 (1 mark)Sol. a = 6, d = 6

Here, t 1

= a = 6

t2

= t1

+ d = 6 + 6 = 12

t3

= t2

+ d = 12 + 6 = 18

t4

= t3

+ d = 18 + 6 = 24

t5

= t4

+ d = 24 + 6 = 30

The first five terms of A.P. are 6, 12, 18, 24 and 30.

NOTE :NOTE :NOTE :NOTE :NOTE : In an A.P. the difference between two consecutive terms is

constant and it is denoted as d and first term of an A.P. is

denoted as a.


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SCHOOL SECTION10

o General term of an A.P. (tn) :

t1

= a = a + 0d t1

= a + (1 1) dt2

= a + d = a + 1d t2

= a + (2 1) dt3

= a + d + d = a + 2d t3

= a + (3 1) dt4

= a + d + d + d = a + 3d t4

= a + (4 1) d

t5 = a + d + d + d + d = a + 4d t5 = a + (5 1) dIn general nth term, of an A.P. t

n= a + (n 1) d


1. Find the twenty fifth term of the A. P. : 12, 16, 20, 24, ..... (2 marks)Sol. For the given A.P. 12, 16, 20, 24, .....

Here, a = t1

= 12d = t

2 t

1= 16 12 = 4

We know,t

n= a + (n 1) d

t25

= a + (25 1) d t

25

= 12 + 24 (4) t

25= 12 + 96

t25

= 108

The twenty fifth term of A.P. is 108.


2. Find the eighteenth term of the A. P. : 1, 7, 13, 19, ..... (2 marks)Sol. For the given A.P. 1, 7, 13, 19, .....

Here, a = t1

= 1d = t

2 t

1= 7 1 = 6

We know,t

n= a + (n 1) d

t18 = a + (18 1) d t

18= 1 + 17 (6)

t18

= 1 + 102 t

18= 103

Eighteenth term of A.P. is 103.

PROBLEM SET - 1 (TEXT BOOK PAGE NO. 162) :

1. Find t11

from the following A.P. 4, 9, 14, ..... . (2 marks)Sol. For the A.P. 4, 9, 14, .....

a = 4, d = 5t

n= a + (n 1) d

t11

= 4 + (11 1) 5

t11 = 4 + 50 t

11= 54


3. Find tnfor an Arithmetic Progression where t

3= 22, t

17= 20. (3 marks)

Given : For an A.P. t3

= 22 and t17

= 20Find : t

n.

Sol. tn

= a + (n 1) dt

3= a + (3 1) d

22 = a + 2d a + 2d = 22 ......(i)

t17

= a + (17 1) d

20 = a + 16d a + 16d = 20 ......(ii)


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SCHOOL SECTION 11

Subtracting (ii) from (i),a + 2d = 22a + 16d = 20

() () (+) 14d = 42

d =42

14 d = 3

Substituting d = 3 in (i),a + 2 ( 3) = 22

a 6 = 22 a = 22 + 6 a = 28

tn

= a + (n 1) d t

n= 28 + (n 1) ( 3)

tn

= 28 3n + 3

tn = 31 3n


4. For an A. P. if t4= 12, and d = 10, then find its general term. (3 marks)

Given : For an A.P. t4

= 12, d = 10Find : General term { t

n}

Sol. tn

= a + (n 1) d t

4= a + (4 1) d

12 = a + 3 ( 10) a = 12 + 30 a = 42

tn = a + (n 1) d t

n= 42 + (n 1) ( 10)

tn

= 42 10n + 10 t

n= 52 10n

The general term of A.P. is 52 10n.


5. Given the following sequence, determine whether it is arithmetic ornot. If it is an Arithmetic Progression, find its general term :

5, 2, 9, 16, 23, 30, ..... (3 marks)Sol. 5, 2, 9, 16, 23, 30, .....

t1

= 5, t2

= 2, t3

= 9, t4

= 16, t5

= 23, t6

= 30t2

t1

= 2 ( 5) = 2 + 5 = 7t3

t2

= 9 2 = 7t4

t3

= 16 9 = 7t5

t4

= 23 16 = 7t6

t5

= 30 23 = 7 The difference between two consecutive terms is 7 which is a constant. The sequence is an A.P. with a = t

1= 5.

Common difference (d) = 7tn

= a + (n 1) dtn

= 5 + (n 1) 7tn

= 5 + 7n 7tn

= 7n 12

The general term of A.P. is 7n 12.


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SCHOOL SECTION12


6. Given the following sequence, determine if it is arithmetic or not. If itis an Arithmetic Progression, find its general term. (2 marks)5, 2, 2, 6, 11, .....

Sol. 5, 2, 2, 6, 11, .....

t1 = 5, t2 = 2, t3 = 2, t4 = 6, t5 = 11t2

t1

= 2 5 = 3t3

t2

= 2 2 = 4t4

t3

= 6 ( 2) = 6 + 2 = 4t5

t4

= 11 ( 6) = 11 + 6 = 5 The difference between two consecutive terms is not a constant. The sequence is not an A.P.


7. How many three digit natural numbers are divisible by 4 ? (3 marks)Sol. The three digit natural numbers that are divisible by 4 are as follows

100, 104, 108, ........ 996.These numbers form an A.P. with a = t1

= 100,d = t

2 t

1= 104 100 = 4.

Let, tn

= 996We know that for an A.P.t

n= a + (n 1) d

996 = 100 + (n 1) 4 996 = 100 + 4n 4 996 = 96 + 4n 4n = 996 96 4n = 900

n =900

4 n = 225

There are 225 three digit natural numbers that are divisible by 4.


7. The sum of first n terms of an A.P. is 3n + n2 ten (i) find first term andsum of first two terms. (ii) find second, third and 15th term. (3 marks)

Sol. Sn

= 3n + n2

S1

= 3(1) + (1)2

S1

= 3 + 1

S1

= 4

S1

= t1

= 4S

2= 3(2) + (2)2

S2

= 6 + 4

S2

= 10

t2

= S2 S

1

t2

= 10 4

t2

= 6

Now,t1+ t

2= 4 + 6

t1+ t

2= 10

a = 4d = 6 4 = 2


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SCHOOL SECTION 13

t3

= t2

+ d t

3= 6 + 2

t3

= 8

tn

= a + (n 1)d

t15 = a + (15 1)d t

15= a + 14d

t15

= 4 + 14(2) t

15= 4 + 28

t15

= 32


8. The 11th term and the 21st term of an A.P. are 16 and 29 respectively, find :(i) the 1st term and the common difference. (2 marks)Sol. t

n= a + (n 1) d

t11

= a + (11 1) d

16 = a + 10d a + 10d = 16 .......(i)

t21

= a + (21 1) d29 = a + 20d

a + 20d = 29 ......(ii)Subtracting (ii) from (i),

a + 10d = 16a + 20d = 29

() () () 10d = 13

10d = 13

d =

13

10 d = 1.3

Substituting d = 1.3 in (i),a + 10 (1.3) = 16

a + 13 = 16 a = 16 13 a = 3

The first term is 3 and the common difference is 1.3

(ii) the 34th term (1 mark)Sol. t

n= a + (n 1) d

t34 = a + (34 1) d= 3 + 33 (1.3)= 3 + 42.9

t34

= 45.9

Thirty fourth term of A.P. is 45.9.

(iii) n such that tn= 55. (1 mark)

Sol. tn

= a + (n 1) d 55 = 3 + (n 1) d 55 = 3 + (n 1) 1.3 55 = 3 + 1.3n 1.3 55 = 1.7 + 1.3n 53.3 = 1.3n


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SCHOOL SECTION14

53.3

1.3= n

n =533

13

n = 41


2. Find first negative term from following A.P. 122, 116, 110, ...... .(Note : find smallest n such that t

n< 0). (4 marks)

Sol. 122, 116, 110, .............a = t

1= 122

d = t2

t1

= 116 122 = 6t

n= a + (n 1) d

tn

= 122 + (n 1) (6)t

n= 122 6n + 6

tn

= 128 6n

when want smallest n such thatt

n< 0

128 6n < 0128 < 6nDividing both sides by 6

128

6< n

21.33... < n

First negative term of A.P. is 4.

oSum of the first n terms of an A.P. :To find sum of first n terms of an A.P. we useGauss technique.

One day Gauss was asked to find sum of all natural

numbers from 1 to 100. He wrote,

S = 1 + 2 + 3 + ..... + 99 + 100

Again he wrote,

S = 100 + 99 + 98 + ..... + 2 + 1

He added both this statements

2S = (100 + 1) + (99 + 2) + (98 + 3) + .....

+ (2 + 99) + (1 + 100)

2S = 101 + 101 + 101 + ..... + 101 (100 times)

2S = 101 100

2S = 10100

S = 5050

The same technique we use for finding sum of first n terms of an A.P. in

the following manner.

The sum of the first n terms of an A.P. denoted by Sn.

If a is the first term and d is the common difference, then the terms of

A.P. are a, a + d, a + 2d, ....... a + (n 1) d, .....

Here, tn

= a + (n 1) d, n N.

Let us denote tn

by l

The first n terms of the A.P. are a, a + d, a + 2d, ......l

2d,l

d,l

Therefore, sum of the first n terms of the A.P. is

But n is term numberwhich is a natural numberThe first natural numbergreater than 21.33... is22

When n = 21t

n= 128 6n

t21 = 128 6 21 t

21= 128 126

t21

= 2When n = 22t

n= 128 6n

t22

= 128 6 22 t

22= 128 132

t22

= 4


15/48


SCHOOL SECTION 15

Sn

= a + (a + d) + (a + 2d) + .... + ( l 2d) + (l d) + l

Also,S

n= l+ (l d) + (l 2d) + .... + (a + 2d) + (a + d) + a

Adding the corresponding terms of the above equations, 2S

n= (a + l) + (a + l) + (a + l) + ...... + (a + l) + (a + l) + (a + l) { n times }

2Sn = n (a + l)

Sn

=n

2(a + l)

Sn

=n

2(t

1+ t

n) .......(i) [ a = t

1and l = t

n]

Here, t1

= a and tn

= [a + (n 1) d]

Sn

=n

2[a + { a + (n 1) d }]

Sn

=n

2[a + a + (n 1) d]

Sn =

n

2 [2a + (n - 1) d] .......(ii)


1. Find the sum of the first n natural numbers and hence find the sum offirst 20 natural numbers. (4 marks)

Sol. The first n natural numbers are 1, 2, 3, 4, ............These natural number form an A.P. with a = 1, d = 1

Sn

=n

2[2a + (n 1) d]

Sn

=n

2[2 (1) + (n 1) 1]

Sn

= n2

[2 + n 1]

Sn

=n

2(n + 1)

S20

=20

2(20 + 1)

S20

= 10 (21) S

20= 210

Sum of first twenty natural numbers is 210.


2. Find the sum of all odd natural numbers from 1 to 150. (4 marks)Sol. The odd natural numbers from 1 to 150 are as follows

1, 3, 5, 7, 9, .........., 149.These numbers form an A.P. with a = 1, d = 2Let, 149 be nth term of an A.P.t

n= 149

tn

= a + (n 1) d149 = 1 + (n 1) 2149 = 1 + 2n 2149 = 2n 1149 + 1 = 2n

2n = 150

n = 75 149 is 75th term of A.P.

Alternative method :

Sn

=n

2[t

1+ t

n]

S20

=20

2[t

1+ t

20]

= 10 [1 + 20]= 10 [21]

S20

= 210


16/48


SCHOOL SECTION16

We have to find sum of 75 terms i.e. S75

Sn

=n

2[2a + (n 1) d]

S75

=75

2[2 (1) + (75 1) 2]

S75

= 752

[2 + 74 [2)]

=75

2[2 + 148]

=75

2(150)

= 75 (75) S

75= 5625

Sum of all odd natural from 1 to 150 is 5625.


3. Find the sum of first 11 positive numbers which are multiples of 6.(2 marks)

Sol. The positive integers which are divisible by 6 are 6, 12, 18, 24, ........The number form an A.P. with a = 6, d = 6.The sum of first 11 positive integers divisible by 6 is (S

11)

Sn

=n

2[2a + (n 1) d]

S11

=11

2[2a + (11 1) d]

=11

2[2 (6) + 10 (6)]

=

11

2 [12 + 60]

=11

2 72

= 11 36S

n= 396

Sum of first 11 positive integers which are divisible by 6 is 396.


3. Find S10

if a = 6 and d = 3. (2 marks)Sol. For an A.P. a = 6, d = 3

Sn =

n

2 [2a + (n 1) d]

S10

=10

2[2a + (10 1) 1]

S10

= 5 [2 (6) + 9 (3)] S

10= 5 (12 + 27)

S10

= 5 (39)

S10

= 195


4. Find the sum of all numbers from 1 to 140 which are divisible by 4.(4 marks)

Sol. The natural numbers from 1 to 140 that are divisible by 4 are as follows :4, 8, 12, 16, .............., 140

Alternative method :

Sn

=n

2[t

1+ t

n]

S75

= 752

[t1

+ t75

]

=75

2[1 + 149]

=75

2[150]

= 75 [75] S

75= 5625


17/48


SCHOOL SECTION 17

These numbers from an A.P. with a = 4, d = t2

t1

= 8 4 = 4Let, 140 be the nth term of A.P.t

n= 140

tn

= a + (n 1) d 140 = 4 + (n 1) 4

140 = 4 + 4n 4 140 = 4n

n =140

4 n = 35 140 is 35 term of A.P. We have to find sum of 35 terms i.e. S

35,

Sn

=n

2[2a + (n 1)d]

S35

=35

2[2 (4) + (35 1) 4]

S35 =

35

2 [8 + 34 (4)]

S35

=35

2[8 + 136)

S35

=35

2[144]

S35

= 2520

Sum of natural numbers from 1 to 140 that are divisible by 4 is 2520.


5. Find the sum of the first n odd natural numbers.Hence find 1 + 3 + 5 + ... + 101. (4 marks)

Sol. The first n odd natural numbers are as follows :1, 3, 5, 7, ............., na = 1, d = t

2 t

1= 3 1 = 2

Sn

=n

2[2a + (n 1)d]

Sn

=n

2[2 (1) + (n 1) 2]

Sn

=n

2[2 + 2n 2]

=n

2

[2n]

Sn

= n2 ......(i)

1 + 3 + 5 + ........ + 101

Let, 101 be the nth term of A.P.

tn

= 101

tn

= a + (n 1) d

101 = a + (n 1) d

101 = 1 + (n 1) 2

101 = 1 + 2n 2

101 = 2n 1

101 + 1 = 2n

2n = 102 n = 51


18/48


SCHOOL SECTION18

101 is the 51st term of A.P., We have to find sum of 51 terms i.e. S

51,

Sn

= n2 [From (i)] S

51= (51)2

S51

= 2601


6. Obtain the sum of the 56 terms of an A. P. whose 19th and 38th terms are52 and 148 respectively. (4 marks)

Sol. t19

= 52, t38

= 148t

n= a + (n 1) d

t19

= a + (19 1) d 52 = a + 18d a + 18d= 52 ......(i)

t38

= a + (38 1) d 148 = a + 37d

a + 37d= 148 ......(ii)Adding (i) and (ii) we get,

a + 18d = 52a + 37d = 148

2a + 55d = 200

Sn

=n

2[2a + (n 1)d]

S56

=56

2[2a + (56 1) d]

S56

=56

2[2a + 55d]

S56

=56

2[200]

S56

= 5600

Sum of first 56 terms of A.P. is 5600.


7. The sum of the first 55 terms of an A. P. is 3300. Find the 28th term.(3 marks)

Sol. S55

= 3300 [Given]

Sn

=n

2

[2a + (n 1)d]

S55

=55

2[2a + (55 1) d]

3300 =55

2[2a + 54d]

3300 =55

22

[a + 27d]

3300

55= a + 27d

300

5 = a + 27d a + 27d = 60 ......(i)


19/48


SCHOOL SECTION 19

tn

= a + (n 1) d t

28= a + (28 1) d

t28

= a + 27d t

28= 60 [From (i)]

Twenty eighth term of A.P. is 60.


8. Find the sum of the first n even natural numbers. Hence find the sumof first 20 even natural numbers. (4 marks)

Sol. The first n even natural numbers are as follows2, 4, 6, 8, ...........

These numbers form an A.P. with a = 2, d = t2

t1

= 4 2 = 2

Sn

=n

2[2a + (n 1)d]

Sn

=n

2

[2 (2) + (n 1) 2]

Sn

=n

2[4 + 2n 2]

Sn

=n

2[2n + 2]

Sn

=n

2 2 (n + 1)

Sn

= n (n + 1) S

20= 20 (20 + 1)

S20

= 20 (21) S

20= 420

Sum of first twenty even natural numbers is 420.


8. For an A.P. given below find t20

and S10

.1 1 1

, , , ......6 4 3

(4 marks)

Sol. For the A.P1 1 1

, , , ......6 4 3

a =1

6

d =1 1

4 6

d =3 2

12 12

d =1

12t

n= a + (n 1)d

t20

=1 1

(20 1)6 12

+

=1 19

6 12+

= 2 1912 12

+


20/48


SCHOOL SECTION20

=21

12

t20

=7

4

Now, Sn =n

2 [2a + (n 1)d]

S10

=10

[2a (n 1)d]2

+

=1 1

5 2 96 12

+

=1 3

53 4

+

=4 9

512

+

= 513

12

S10

=65

12


5. From an A.P. first and last term is 13 and 216 respectively. Commondifference is 7. How many terms are there in that A.P. Find the sum ofall terms. (4 marks)

Sol. for an A.P a = 13

Let last term be tn = 216d = 7t

n= a + (n 1)d

tn

= 13 + (n 1)7 216 = 13 + 7n 7 216 = 6 + 7n 216 6 =7n 210 = 7n n = 30

Sn

=n

2[2a + (n 1)d]

S30 = 302[2a + (30 1)d]

S30

= 15[2(13) + 29 (7)] S

30= 15 [26 + 203]= 15 (229)

S30

= 3435

Sum of all terms of AP is 3435.


6. Second and fourth term of on A.P. is 12 and 20 respectively. Find thesum of first 25 terms of that A.P. (4 marks)

Sol. t2 = 12, t 4 = 20tn

= a + (n 1)d


21/48


SCHOOL SECTION 21

t2

= a + (2 1)d12 = a + d

a +d = 12 .......(i)t

4= a + (4 1)d

20 = a + 3d

a + 3d= 20 .......(ii)Subtracting (ii) from (i),a + d = 12a + 3d = 20() () ()

2d = 8 d = 4

Substituting d = 4 in (i),a + 4 = 12

a = 12 4 a = 8

Sn = n2[2a + (n 1)d]

S25

=25

2[2a + (25 1)d]

=25

2[2(8) + 24 (4)]

=25

2[16 + 96]

=25

2[112]

S25 = 1400

Sum of 25 terms of the A.P is 1400.


4. In the A.P. 7, 14, 21, ....... How many terms are have to consider forgetting sum 5740. (4 marks)

Sol. for the A.P. 7, 14, 21a = 7, d = 7

Sn

= 5740

Sn

=n

2[2a + (n 1) d]

Sn

=n

2[ 2(7) + (n 1)7]

5740 =n

2[ 14 + 7n 7]

5740 =n

2[ 7n + 7]

11480 = 7n2 + 7n 7n2 + 7n 11480 = 0

Dividing through at by 7 we get,n2 + n 1640 = 0

n2 + 41n 40n 1640 = 0

n (n + 41) 40 (n + 41) = 0 (n + 41) (n 40) = 0


22/48


SCHOOL SECTION22

n + 41 = 0 or n 40 = 0 n = 41 or n = 40

n = 41 is not acceptable because no of terms cannot be negative n = 40

For the given sequence 40 terms have to be considered for gettingsum of 5740.

o Properties of an A.P. :

- - - - -

For an A.P. with first term a and the common difference d, if any real numberk is added to each term of an A.P. then the new sequence is also an A.P. withthe first term. a + k and the same common difference d.For example : For an A.P. 5, 8, 11, 14, ......a = 5, d = 3

Add any real number k to each term of the A.P. we get

5 + k, 8 + k, 11 + k, 14 + k ......t1

= 5 + kt2

t1

= 8 + k (5 + k) = 8 + k 5 k = 3t3

t2

= 11 + k (8 + k) = 11 + k 8 k = 3t4

t3

= 14 + k (11 + k) = 14 + k 11 k = 3

- - - - -

For an A.P. with the first term a and the common difference d, if each termof an A.P. is multiplied by any real number k, then the new sequence is alsoan A.P. with the first term ak and the common difference dk.For example : For an A.P. 5, 8, 11, 14, ......

Multiplying all the terms by any real number k we get,5k, 8k, 11k, 14k ......t1

= 5kt2

t1

= 8k 5k = 3kt3

t2

= 11k 8k = 3kt4

t3

= 14k 11k = 3k

We assume three, four or five consecutive terms in an A.P. in thefollowing manner :

(i) Three consecutive terms as a d, a, a + d(ii) Four consecutive terms as a 3d, a d, a + d, a + 3d

(here note that the common difference is 2d)(ii i) Five consecutive terms as a 2d, a d, a, a + d, a + 2d


1. Find four consecutive terms in an A.P. whose sum is 12 and the sum of3rd and 4th term is 14. (4 marks)

Sol. Let the four consecutive terms in an A.P. bea 3d, a d, a + d, a + 3d

As per the first condition,a 3d + a d + a + d + a + 3d = 12

4a = 12 a = 3

As per the second condition,a + d + a + 3d = 14

2a + 4d = 14 2 (3) + 4d = 14


23/48


SCHOOL SECTION 23

6 + 4d = 14 4d = 14 6 4d = 8 d = 2 a 3d = 3 3 (2) = 3 6 = 3

a d = 3 2 = 1a + d = 3 + 2 = 5a + 3d = 3 + 3 (2) = 9

The four consecutive terms of A.P. are 3, 1, 5 and 9.


2. Find four consecutive terms in an A.P. whose sum is 54 and the sum of1st and 3rd term is 30. (4 marks)

Sol. Let the four consecutive terms is an A.P. be a 3d, a d, a + d and a + 3dAs per first condition,a 3d + a d + a + d + a + 3d = 54

4a = 54

a =54

4

a =27

2

As per the second condition,a 3d + a + d = 30

2a 2d = 30

227

2

2d = 30

27 2d = 30 2d = 30 + 27 2d = 3

d =3

2

a 3d =27

2

33

2

= 27 9

2=

36

2= 18

a d =27 3

2 2

=30

2= 15

a + d =27 3

2 2

+ =27 3

2 2

+ =24

2= 12

a + 3d =27 3 32 2

+

= 27 9

2+

=27 9

2+

=18

2= 9

The four consecutive terms of an A.P. are 18, 15, 12 and 9.


3. Find three consecutive terms in an A.P. whose sum is 3 and the productof their cubes is 512. (4 marks)

Sol. Let three consecutive terms in an A.P. bea d, a, a + d

As per the first given condition,a d + a + a + d = 3

3a = 3 a = 1


24/48


SCHOOL SECTION24

As per the second given condition,(a d)3 a3 (a + d)3 = 512

[(a d) a (a + d)]3 = 512Taking cube roots on both sides

(a d) a (a + d) = 5123

(a d) a (a + d) = 8 a (a d) (a + d) = 8 a (a2 d2) = 8 1 [( 1)2 d2] = 8 1 (1 d2) = 8 d2 1 = 8 d2 = 8 + 1 d2 = 9 d = +3

If d = 3 a d = 1 3 = 4 a d = 1(3) = 2 a + d = 1 + 3 = 2 a + d = 1 3 = 4

The three consecutive terms of A.P. are 4, 1, 2 or 2, 1, 4


4. In winter, the temperature at a hill station from Monday to Friday is inA.P. The sum of the temperatures of Monday, Tuesday and Wednesdayis zero and the sum of the temperatures of Thursday and Friday is 15.Find the temperature of each of the five days. (4 marks)

Sol. Let the temperatures of hill station from Monday to Friday which formare A.P. bea 2d, a d, a, a + d, a + 2d respectively.

As per the first condition,

a 2d + a d + a = 0 3a 3d = 0 3 (a d) = 0 a d = 0 a = d

As per the second condition,a + d + a + 2d = 15

2a + 3d = 15 2a + 3a = 15 [ a = d] 5a = 15 a = 3 d = 3 [ d = a]

a 2d = 3 2 (3) = 3 6 = 3 a d = 3 3 = 0 a + d = 3 + 3 = 6 a + 2d = 3 + 2 (3) = 3 + 6 = 9

The temperatures from Monday to Friday are 3, 0, 3, 6 and 9 respectively.

o WORD PROBLEMS FOR A.P.


1. Mary got a job with a starting salary of Rs. 15000/- per month. She willget an incentive of Rs. 100/- per month. What will be her salary after20 months? (4 marks)

Sol. Since Marys salary increases by Rs. 100 every month the successivesalaries are in A.P.


25/48


SCHOOL SECTION 25

Starting salary of Marry (a) = Rs. 15000Monthly incentive in salary (d) = 100No. of months (n) = 20Salary after twenty months = t

20= ?

tn

= a + (n 1) d

t20 = a + (20 1) d t

20= 15000 + 19 (100)

t20

= 15000 + 1900 t

20= 16900

Marry salary after twenty months is Rs. 16900.


2. The taxi fare is Rs. 14 for the first kilometer and Rs. 2 for each additionalkilometer. What will be fare for 10 kilometers ? (3 marks)

Sol. Since the taxi fare increases by Rs. 2 every kilometer after thefirst, the successive taxi fares form an A.P.

The taxi fare for first kilometer (a) = Rs. 14Increase in taxi fare in every kilometer after first kilometer (d) = 2No. of kilometers covered by taxi (n) = 10

Taxi fare for 10 kilometers = t10

= ?t

n= a + (n + 1) d

t10

= a + (10 1) d t

10= 14 + 9 (2)

t10

= 14 + 18 t

10= 32

Taxi fare for ten kilometers is Rs. 32.


3. Mangala started doing physical exercise 10 minutes for the first day.She will increase the time of exercise by 5 minutes per day, till shereaches 45 minutes. How many days are required to reach 45 minutes ?

(3 marks)

Sol. Since the workout time Mangala increases by 5 minutes everydayafter the first day, the successive workout times are in A.P.

Workout time for first day (a) = 10 minutes.Increases in workout time (d) = 5 minutesLet No. of days required to reach workout time of 45 minutes be n days.t

n= 45

tn

= a + (n 1) d 45 = 10 + (n 1) 5 45 = 10 + 5n 5 45 = 5 + 5n 45 5 = 5n 5n = 40 n = 8

8 days required to reach work out time of 45 minutes.


4. There is an auditorium with 35 rows of seats. There are 20 seats in thefirst row, 22 seats in the second row, 24 seats in the third row, and soon. Find the number of seats in the twenty fifth row. (3 marks)

Sol. Since the no. of seats in each row of the auditorium are 20, 22, 24, ......The no. of seats in each row form an A.P.


26/48


SCHOOL SECTION26

No. of seats in first row (a) = 20Difference in no. of seats in two successive rows is (d) = 2No. of seats in 25th row = t

25= ?

tn

= a + (n + 1) d t

25= a + (25 1) d

t25 = 20 + 24 (2) t

25= 20 + 48

t25

= 68

There are 68 seats in 25th row.


5. A village has 4000 literate people in the year 2010 and this numberincreases by 400 per year. How many literate people will be there till theyear 2020 ? Find a formula to know the number of literate people aftern years ? (4 marks)

Sol. Since the no. of literate people increases by 400 every year, the

population of literate people every year forms an A.P.No. of people in the year 2010 (a) = 4000Increase in population every year (d) = 400No. of years from 2010-2020 (n) = 10t

n= a + (n 1) d

t10

= 4000 + (10 1) 400 t

10= 4000 + 9 (400)

t10

= 4000 + 3600 t

10= 7600

There will be 7600 literate people till the year 2020t

n= a + (n 1) d

tn = 4000 + (n 1) 400

tn

= 4000 + 400n 400 t

n= 3600 + 400n

There will be (3600 + 400n) literate people after n years.


6. Neela saves in a Mahila Bachat gat Rs. 2 on the first day, Rs.4 on thesecond day, Rs. 6 on the third day and so on. What will be her saving inthe month of February 2010 ? (4 marks)

Sol. The savings done by Neela on each day is as follows 2, 4, 6, .......These every day savings form an A.P. with

First day saving (a) = 2Difference in savings made in two successive days (d) = 2Total no. of days in the month of February 2010 (n) = 28

Total savings for the month of February (S28

) = ?

Sn

=n

2[2a + (n 1) d]

S28

=28

2[2 (2) + (28 1) (2)]

= 14 [4 + 27 (2)]= 14 [4 + 54]

S28

= 14 [58] S

28

= 812

Neela saved Rs. 812 in the month of February.


27/48


SCHOOL SECTION 27


7. Babubhai borrows Rs. 4000 and agrees to repay with a total interest ofRs. 500. in 10 instalments, each instalment being less that the precedinginstalment by Rs. 10. What should be the first and the last instalment?

(4 marks)

Sol. Total money repaid by Babubhai in 10 instalments = (S10)= 4000 + 500= Rs. 4500

No. of instalments (n) = 10Difference between two consecutive instalments (d) = 10First instalment = (a) = ?Last instalment (t

10) = ?

Sn

=n

2[2a + (n 1) d]

S10

=10

2[2a + (10 1) d]

4500 = 5 [2a + 9 ( 10)]

4500

5= 2a 90

900 = 2a 90 900 + 90 = 2a 990 = 2a

990

2= a

a = 495t

n= a + (n 1) d

t10

= a + (10 1) d

t10 = 495 + 9 ( 10) t

10= 495 90

t10

= 405

First instalment is Rs. 495 and last instalment is Rs.405.


8. A meeting hall has 20 seats in the first row, 24 seats in the second row,28 seats in the third row, and so on and has in all 30 rows. How manyseats are there in the meeting hall ? (4 marks)

Sol. The no. of seats in each row are as follows 20, 24, 28, .........The no. of seats in each row form an A.P. with

First term (a) = 20Difference between the no. of seats in two successive rows (d) = 4

Total no. of rows (n) = 30Total no. of seats in 30 rows (S

30) = ?

Sn

=n

2[2a + (n 1) d]

S30

=30

2[2 (20) + (30 1) 4]

S30

= 15 [40 + 116] S

30= 15 (156)

S30

= 2340

Total no. of seats in the meeting hall is 2340.


28/48


SCHOOL SECTION28


9. Vijay invests some amount in National saving certificate. For the 1st yearhe invests Rs. 500, for the 2nd year he invests Rs. 700, for the 3rd yearhe invests Rs. 900, and so on. How much amount he has invested in12 years ? (4 marks)

Sol. Yearly investments of Vijay are as follows 500, 700, 900, .......The yearly investments form an A.P. with first year investment (a) = 500Difference between investment done in two successive years (d) = 200.No. of years (n) = 12

Total investment done in 12 years = (S12

) = ?

Sn

=n

2[2a + (n 1) d]

S12

=12

2[2 (500) + (12 1) 200]

S12

= 6 [1000 + 11 (200)] S

12= 6 [1000 + 2200]

S12

= 6 [3200] S12 = 19200

Total investment done in 12 years is Rs. 19200.


10. In a school, a plantation program was arranged on the occasionof world environment day, on a ground of triangular shape.

The trees are to be planted as shown in the figure. One plantin the first row, two in the second row, three in the third rowand so on. If there are 25 rows then find the total number ofplants to be planted. (4 marks)

Sol. The number of trees in each row upto the 25th row areas follows :

1, 2, 3, 4, ...........These no. of trees planted in each row forms an A.P. withNo. of trees in first row (a) = 1Difference between no. of trees planted in two successive rows (d) = 1No. of rows (n) = 25

Sn

=n

2[2a + (n 1) d]

S25

=25

2[2 (1) + (25 1) 1]

S25

=25

2[2 + 24]

S25 = 252 (26)

S25

= 25 (13) S

25= 325

325 trees were planted in 25 rows.

o GEOMETRIC PROGRESSION :Geometric progression is a sequence such that the given first term each termso obtained by multiplying a non-zero constant r to the preceeding term.

Also it is a sequence in which the ratio of to two consecutive terms in theprogression is constant.Consider, 2, 4, 8, 16, ......

We have t 4t 2=

2

1= 2, t 8 2t 4

= =3

2, t 16 2t 8

= =4

3and so on.


29/48


SCHOOL SECTION 29

Here for a G.P. first term is denoted as a and common ratio is denoted as r.General term of a G.P.

Thus in general any G. P. can be expressed as a, ar, ar2, ar3, a is firstterm and r is the common ratio.

The nth term of a G.P.

Consider a G.P. whose first term is a and the common ratio is rt1, t

2, t

3, t

4, .......

Then, t1

= a, t2

= ar, t3

= ar2, t4

= ar3

t

t2

1= r,

t

t3

2= r, .......,

t

tn

n1= r

Multiplying all (n 1) ratios we get,t t t t

......t t t t

2 3 4 n

1 2 3 n1= r r r r r r ..... (n 1) times

t

tn

1= r(n1) i.e.

t

an

= r(n1), tn

= ar(n1)

Hence in general tn = arn1

Thus the nth term of a G.P. with the first term a and the common ratio r is t

n= arn1


1. Find the ninth term of the G.P. 3, 6, 12, 24, .... (3 marks)Sol. For the G.P.

3, 6, 12, 24, ........a = 3

r =t

t2

1

=6

3= 2

tn

= arn 1

t9

= 3 (2)9 1

t9

= 3 (2)8

t9

= 3 (256) t

9= 768

The ninth term of the G.P. is 768.


2. Write down the first five terms of the geometric progression which hasfirst term 1 and common ratio 4. (2 marks)

Sol. For the G.P.The first term (a) = 1Common ratio (r) = 4

t1 = a = 1t

2= ar = 1 4 = 4

t3

= ar2 = 1 (4)2 = 16t

4= ar3 = 1 (4)3 = 64

t5

= ar4 = 1 (4)4 = 256

The first five terms of the G.P. are 1, 4, 16, 64 and 256.


3. Find the 4th and 9th terms of the G.P. with first term 4 and common ratio2. (3 marks)

Sol. For the G.P.

The first term (a) = 4Common ratio (r) = 2


30/48


SCHOOL SECTION30

tn

= arn 1

t4

= ar4 1

t4

= ar3

t4

= 4 (2)3

t4

= 4 (8)

t4 = 32t

9= ar9 1

t9

= ar8

t9

= 4 (2)8

t9

= 1024

The fourth and ninth term of the G.P. are 32 and 1024 respectively.


4. Find the common ratio and the 7th term of the G.P. 2, 6, 18, .... (2 marks)Sol. For the G.P. 2, 6, 18, ........

The first term (a) = 2

Common ratio (r) =6

2 = 3Seventh term t

7= ?

tn

= arn 1

t7

= ar7 1

t7

= ar6

t7

= 3 (3)6

t7

= 2 (729) t

7= 1458

The common ratio of the G.P. is 3 and the seventh term is 1458.


5. Find the 69th term of the G.P. 1, 1, 1, 1, .... (2 marks)Sol. For the G.P. 1, 1, 1, 1

First term (a) = 1

Common ratio (r) =1

1= 1

tn

= arn 1

t69

= ar69 1

t69

= ar68

t69

= 1 ( 1)68

t69

= 1 1 t

69= 1

Sixtyninth term of G.P. is 1.


6. Find the 15th term of the G.P. 3, 12, 48, 192, ... (2 marks)Sol. For the G.P. 3, 12, 48, 192, ........

First term (a) = 3

Common ratio (r) =12

3= 4

tn

= arn 1

t15

= ar15 1

t15

= ar14

t15

= 3 (4)14

t15 = 3 4

14

Fifteenth term of G.P. is 3 414.


31/48


SCHOOL SECTION 31

o Sum of first n terms of a G.P. :

Suppose that we want to find the sum of the first n terms of a geometricprogression a, ar, ar2, ar3 ...... where r 1.Let, S

n= a + ar + ar2 + ar3 + ..... + ar(n1) ......(i)

Multiplying the above equation by r we get,

rSn = ar + ar2 + ar3 + ..... + ar(n1) + arn .....(ii)

Subtracting equation (ii) from equation (i) we get,S

n rS

n= a arn

So that, Sn

(1 r) = a (1 rn)Dividing by (1 r) (since r 1) we get,

Sn

=a (1 r )

1 r

n

[We use this if r < 1]

This can also be written as

Sn

=a (r 1)

r 1

n

[We use this if r > 1]

NOTE :NOTE :NOTE :NOTE :NOTE : When the common ratio r = 1, the G.P. becomes a, a, a, a, .....In this case clearly, the sum of the n terms is a + a + a + .....,n times = na.


1. Find the indicated sums for the following Geometric Progressions.

(i) 2, 6, 18, ... Find S7. (2 marks)

Sol. 2, 6, 18, .........a = 2

r =6

2= 3

Sn

=a (1 r )

1 r

n

S7

=a (1 r )

1 r

7

S7

=2 (1 3 )

1 3

7

S7 =

2 (1 3 )

2

7

S7

= 1 (1 37) S

7= 37 1

S7

= 2187 1

S7

= 2186

(ii) 2, 4, 8, 16, ... Find S9

and S12

. (3 marks)

Sol. For the G.P. 2, 4, 8, 16, .......

a = 2

r =

4

2= 2

Sn

=a (1 r )

1 r

n

S9

=a (1 r )

1 r

9

S9

=2 (1 ( 2) )

1 ( 2)

9

S9

=2 (1 ( 512))

1 2+

S9

=2 (1 512)

3

+

S9

=1026

3

S9

= 342

S12

=a (1 r )

1 r

12

S12

=2 (1 ( 2) )

1 ( 2)

12

S12

=2 (1 4096)

3

S12

=2 ( 4095)

3

S12

= 2 ( 1365)

S12

= 2730


32/48


SCHOOL SECTION32

(iii) 1,1

2, 2

1

2, 3

1

2, ... Find S

6.

(2 marks)

Sol. a = 1

r =121

=1

2

Sn

=a (1 r )

1 r

n

S6

=

11 1

2

11

2

6

S6

=

1

1 641

2

S6

=64 1 1

64 2

S6

=63 2

64 1

S6

=63

32

(iv) 1, 2 , 2, ... Find S10. (3 marks)

Sol. For the G.P.a = 5

r = 2

Sn

=a (1 r )

1 r

n

S10

=( )a 1 r 1 r

10

S10

=( )1 1 2

1 2

10

S10

=

1 2

1 2

101

2

S10

=1 2

1 2

5

S10

=1 32

1 2

S10

=31

1 2

S10

=( )

( ) ( )

31 1 2

1 2 1 2

+

+

S10

=

( )

( )

31 1 2

1 2

+

22

S10

=( ) 31 1 2

1 2

+

S10

=( ) 31 1 21

+

S10

= ( )31 1 2+


2. If in a G.P. r= 2 and t8

= 64 then find aand S6. (3 marks)

Sol. For a G.P.r = 2t

8= 64

tn

= arn 1

t8

= ar8 1

64 = ar 7

64 = a (2)7

64 = a (128)

64128

= a


33/48


SCHOOL SECTION 33

a =1

2

Sn

=a (1 r )

1 r

n

S6

= ( )a 1 r 1 r

6

S6

=

1(1 2 )

21 2

6

S6

=

1(1 64)

21

S6

=

1( 63)

21

S6

= 632

a =1

2and S

6=

63

2.


3. If S3

= 31 and S6

= 3906 then find aand r. (4 marks)Sol. For a G.P.

Sn

=a (1 r )

1 r

n

S3

= a (1 r )1 r

3

But,S

3= 31 [Given]

a (1 r )

1 r

3

= 31 .......(i)

Similarly,

S6

=a (1 r )

1 r

6

But,S

6= 3906 [Given]

a (1 r )

1 r

6

= 3906 ......(ii)

Dividing (ii) by (i),

a (1 r ) 1 r

1 r a (1 r )

6

3 =3906

31

1 r

1 r

6

3 = 126

1 (r )

1 r

3 2

3 = 126

(1 r ) (1 r )1 r

+3 3

3 = 126


34/48


SCHOOL SECTION34

1 + r3 = 126 r3 = 126 1 r3 = 125

Taking cube roots on both sides,r = 5

Substituting r = 5 in (i),a (1 5 )

1 5

3

= 31

a (1 125)

4 = 31

a ( 124) = 31 4 124a = 124

a =124

124

a = 1 a = 1 and r = 5.


4. If S6

= 126 and S3

= 14 then find aand r. (4 marks)Sol. For a G.P.

Sn

=a (1 r )

1 r

n

S6

=a (1 r )

1 r

6

But, S6= 126 [Given]

a (1 r )

1 r

6

= 126 .....(i)Similarly,

S3

=a (1 r )

1 r

3

But, S3

= 14 [Given]

a (1 r )

1 r

3

= 14 .....(ii)

Dividing (i) by (ii),

a (1 r ) a (1 r )

1 r 1 r

6 3

=126

14

a (1 r ) 1 r

1 r a (1 r )

6

3 = 9

1 r

1 r

6

3 = 9

1 (r )

1 r

2 3 2

3 = 9

(1 r ) (1 r )

1 r

+3 3

3 = 9

1 + r3 = 9 r3 = 9 1 r3 = 8

Taking cube roots on both sidesr = 2


35/48


SCHOOL SECTION 35

Substituting r = 2 in (ii), we get,

a (1 2 )

1 2

3

= 14

a (1 8)

1 = 14

a ( 7)

1 = 14

7a = 14 a = 2

a = 2, r = 2.


5. If the nth, (2n)th, (3n)th terms of a G.P. are a, b, crespectively then showthat b2 = ac. (4 marks)

Sol. For the G.P.Let first term be A

Common ratio be r t

n= Arn 1

But,tn

= a [Given] Arn 1 = a ......(i)

t2n

= Ar2n 1

But,t

2n= b [Given]

Ar2n 1 = b ......(ii)t

3n= Ar3 1

But,t

3n= c [Given]

Ar3n 1 = c .....(iii)b = Ar2n 1 [From (ii)]

Squaring both sides,b2 = (Ar 2n 1)2

b2 = A2 (r2n 1)2

b2 = A2 r4n 2 ......(iv)Multiplying (i) and (iii) we get,

ac = Ar n 1 . Ar3n 1

ac = A 2 rn 1+ 3n 1

ac = A 2 r4n 2 .....(v) From (iv) and (v) we get, b2 = ac


9. In a school, tree plantation on Independence day was arranged. Everystudent from I standard will plant 2 trees, II standard students will plant4 trees each, III standard students will plant 8 trees each etc. If there are5 standard, how many trees are planted by the student of that school?

(4 marks)

Sol. No of trees planted by a student from 1st, 2nd, 3rdstandards are 2, 4,8 respectively

The no. of trees planted by each student form a G.P with

a = 2 , r =4

2= 2

There are 5 standards. Total no. of trees planted by the student of that school is S

5


36/48


SCHOOL SECTION36

Sn

=a (1 r )

1 r

n

S5

=( )2 1 (x)1 2

5

S5

=2 (1 32)

1 2

S5

=2 (31)

1

S5

= 62

Total no of trees planted by student of school is 62 trees.

o ARITHMETIC MEAN :

If three numbers x, y, z are in A.P. then y is called the Arithmetic mean

between x and z.To find the Arithmetic mean between any two numbers x and y.

Suppose x and y are any two numbers. Let the Arithmetic mean between x

and y be A. Then x, A, y are in A.P.

A x = y A 2A = x + y

A =x y

2

+

Thus the arithmetic mean A between any two numbers x and y is given by

A =x y

2

+

o GEOMETRIC MEAN :

If three numbers x, y, z are in G.P. then y is called the geometric meanbetween x and z.The find the geometric mean between any two numbers x and y.Suppose x and y are any two numbers with the same sign. Let the Geometricmean between x and y be G, Then x, G, y are in G.P.

G

x=

y

G

G2 = xy

G = xy

Thus Geometric mean G between any two numbers x and y is given by

G = xy


1. Find three consecutive terms in a G.P. such that the sum of the firsttwo terms is 9 and the product of all the three is 216. (4 marks)

Sol. Let three consecutive terms of G.P. bea

r, a, ar

As per first given condition,

ar

+ a = 9 ......(i)


37/48


SCHOOL SECTION 37

As per second condition,

a

r a ar = 216

a3 = 216Taking cube roots on both side,

a = 2163

a = 6Substituting a = 6 in (i),

6

r+ 6 = 9

6 6r

r

+= 9

6 + 6r = 9r 6 = 9r 6r 3r = 6 r = 2

a

r=

6

2= 3

ar = 6 2 = 12

The three consecutive terms of G.P. are 3, 6, 12.


2. Find three consecutive terms in a G.P. such that the sum of the 2ndand 3rd term is 60 and the product of all the three is 8000. (4 marks)

Sol. Let three consecutive terms of G.P. bea

r, a, ar.

As per first condition,a + ar = 60 ......(i)

As per second condition,

a

r a ar = 8000

a3 = 8000Taking cube roots on both side,

a = 80003

a = 20 20 20 3

a = 20

Substituting a = 20 in (i),20 + 20r = 60

20r = 60 20 20r = 60 20 20r = 40

r =40

20 r = 2

a

r=

20

2= 10

ar = 20 2 . ar = 40

The three consecutive terms of the G.P. are 10, 20, 40.


38/48


SCHOOL SECTION38


3. Sachin, Sehwag and Dhoni together scored 228 runs. Their individualscores are in G.P. Sehwag and Dhoni together scored 12 runs more thanSachin. Find their Individual scores. (5 marks)

Sol. Let individual scores of Sachin, Sehwag and Dhoni be

a

r, a and arAs per the first given condition,a

r+ a + ar = 228 .....(i)

As per second given condition,

a + ar = 12 +a

r......(ii)

Substituting (ii) in (i),a

r+ 12 +

a

r= 228

12 +2a

r= 228

2a

r= 228 12

2a

r= 216

a

r=

216

2

a

r= 108 .....(iii)

a = 108r Substituting (iii) in (i),

108r

r + 108r + (108 r)r = 228a

108 , a 108r r

= =

Multiplying throughout by r,

108 + 108r + 108r2 = 228

108r2 + 108r + 108 228 = 0

108r2 + 108r 120 = 0

Dividing throughout by 12 we get,

9r2 + 9r 10 = 0

9r2 + 15r 6r 10 = 0

3r (3r 5) 2 (3r 5) = 0

(3r 5) (3r 2) = 0

3r 5 = 0 or 3r 2 = 0

3r = 5 or 3r = 2

r =5

3or r =

2

3

If r =5

3If r =

2

3a = 108r a = 108r

a = 108 5

3a = 108

2

3a = 36 5 a = 36 2a = 180 a = 72

ar = 180 5

3ar = 72

2

3 ar = 300 ar = 48


39/48


SCHOOL SECTION 39

r =5

3is not acceptable because ar (individual score) cannot be 300

as the sum of the scores is 228. ar = 48

The runs scored by Sachin, Sehwag and Dhoni are 108 runs,

72 runs and 48 runs.


4. If 25 is the arithmetic mean between x and 46, then find x. (2 marks)Sol. 25 is the arithmetic mean between x and 46

25 =x 46

2

+

50 = x + 46 x = 50 46

x = 4


5. If x + 3 is the geometric mean between x + 1 and x + 6 then find x. (2 marks)Sol. x + 3 is the geometric mean between x + 1 and x + 6.

(x + 3)2 = (x + 1) (x + 6) x2 + 6x + 9 = x 2 + 6x + x + 6 9 = x + 6 9 6 = x

x = 3


6. Find the geometric mean of 82 1 and 82 +1. (2 marks)

Sol. Let G be the geometric mean of 82 1 and 82 1+

G2 = ( ) ( )82 1 82 1+

G2 = ( )822

12

G2 = 82 1 G2 = 81

Taking square roots on both sides,

G = 81 G = +9

The geometric mean of 81 1 and 81 1+ is 9 or 9


7. If the arithmetic mean and the geometric mean of two numbers are inthe ratio 5 : 4 and the sum of the two numbers is 30 then find thesenumbers. (4 marks)

Sol. Let the two numbers be x and yLet Arithmetic mean of x and y be denoted as A and geometricmean of x and y be G.

A =x y

2

+......(i)

G = + xy .....(ii) x + y = 30 .....(iii) [Given]


40/48


SCHOOL SECTION40

Substituting (iii) in (i),

A =30

2 A = 15

G =

5

4 [Given]

15

G=

5

4

4

155

= G

G = 12

xy = 12

Squaring both sides,xy = 144

y =144

x

......(iv)

Substituting (iv) in (iii), we get,

x +144

x= 30

Multiplying throughout by x we get,x2 + 144 = 30x

x2 30x + 144 = 0 x2 24x 6x + 144 = 0 x (x 24) 6 (x 24) = 0 (x 24) (x 6) = 0 x 24 = 0 or x 6 = 0 x = 24 or x = 6

If x = 24 If x = 6

y =144

xy =

144

x

y =144

24y =

144

6 y = 6 y = 24

The two numbers are 6 and 24.

MCQs

1. Select the correct sequence of that numbers satisfies for an G.P.

(a) 1, 3, 6, 10, .... (b) 3, 6, 12, 24(c) 10, 17, 16, 19 (d) 22, 26, 28, 31

2. Select the correct sequence of that numbers not satisfies for an A.P.(a) 0.5, 2, 3.5, 5 (b) 22, 26, 28, 31(c) 3, 5, 7, 9, 11 (d) 1, 4, 7, 10

3. For an A.P. 4, 9, 14, ............. then t11

= ............. .(a) 49 (b) 54(c) 59 (d) 44

4. The 18th term of an A.P. 1, 7, 13, 19, ............. is(a) 103 (b) 109(c) 97 (d) 115


41/48


SCHOOL SECTION 41

5. The general term of an A.P. is tn

............. .(a) a (n 1) d (b) a + (n 1) d(c) a (n + 1) d (d) a + (n + 1) d

6. Sum of 1st n terms of an A.P. Sn

= ............. .

(a ) n [a (n 1)d]2

+ (b) n [2a (n 1)d]2

+ +

(c)n

[2a (n 1)d]2

+ (d)n

[a (n 1)d]2

+ +

7. If a = 6 and d = 3. S10

= ?(a) 192 (b) 195(c) 198 (d) 201

8. The general form of an G.P. tn

= ............. .

(a )a

rn 1 (b)

r

a

n 1

(c) ar n 1 (d) ra n1

9. The 9th term of an G.P. 3, 6, 12, 24 is ............. .(a) 384 (b) 768(c) 1536 (d) 192

10. The 7th term of G.P. 2, 6, 18 is ............. .(a) 1458 (b) 1458(c) 486 (d) 486

11. Sum for 1st in terms of G.P. for r > 1 is Sn

= ............. .

(a )a (r 1)

r 1

n

(b)a (r )

1 r

n 1

(c)a (r 1)

r 1

+

+

n

(d)a (r 1)

rn 1

12. Three consecutive numbers are an G.P. and their product is 1000 then thesecond term a is ............. .

(a )1

10(b) 1

(c) 10 (d) 100

13. The 9th term of an G.P. 3, 6, 12, 24, ..... is ............. .(a) 3(2)8 (b) 3(2)9

(c)1

3 2

8

(d)1

32

9

14. If (x + 3) is geometric mean of (x + 1) and (x + 6) then x = ............. .(a) 9 (b) 6(c) 15 (d) 3

15. If 25 is arithmetic mean of x and x + 46 then x = ............. .

(a) 36 (b) 2(c) 46 (d) 50


42/48


SCHOOL SECTION42

16. Arithmetic mean A between any two number is given by A = ............. .

(a )x y

2(b)

x y

2

+

(c) 2 (x + y) (d)2

x y+

17. If G is the geometric mean (GM) of two numbers x, y then ............. .(a) G2 = xy (b) G = xy

(c)x y

G2

+= (d)

xyG

2=

18. Three consecutive numbers are in A.P. such that the sum of the first andthe last is 8, so that the second term a is ............. .(a) 4 (b) 2(c) 0 (d) 2

19. The sum of first 10 natural numbers is ............. .(a) 55 (b) 155(c) 310 (d) 210

20. In G.P. 4 consecutive term is ............. .

(a )a a

, , ar, ar rr

2

2 (b)a a

, , ar, ar rr

3

3

(c)a a

, ar, +ar,r r

(d)a a a

, , + , arr rr

+3

3

: ANSWERS :

1. (b) 3, 6, 12, 24 2. (b) 22, 26, 28, 313. (b) 54 4. (a) 103

5. (b) a + (n 1) d 6. (c)n

[2a (n 1)d]2

+

7. (b) 195 8. (c) ar n 1

9. (b) 768 10. (b) 1458

11. (a)a (r 1)

r 1

n

12. (c) 10

13. (a) 3 (2)8 14. (d) 3

15. (b) 2 16. (b)x y

2

+

17. (a) G2

= xy 18. (a) 419. (a) 55 20. (b)

a a, , ar, ar

rr

3

3

Open End Questions

1. Write first four terms of a G.P. whose first term is 3.Sol. For a G.P.

a = 3Let us take r = 2t

1= a = 3

t2 = ar = 3 2 = 6t

3= ar2 = 3 2 2 = 12


43/48


SCHOOL SECTION 43

t4

= ar3 = 3 2 2 2 = 24

The first four terms of G.P. one of the possibility is 3, 6, 12, 24.

2. Write first four terms of a G.P. whose common ratio is 2.Sol. For a G.P.

r = 2Let us take a = 2t

1= a = 1

t2

= ar = 2 ( 2) = 4t

3= ar2 = 2 ( 2)2 = 2 4 = 8

t4

= ar3 = 2 ( 2)3 = 2 ( 8) = 16

One of the possible ways of writing the first form terms of G.P.could be 2, 4, 8 and 16.

3. Write first four terms of an A.P. whose first term is 10.Sol. For an A.P. a = 10

Let us take d = 5t

1= a = 10

t2

= a + d = 10 + 5 = 15t

3= a + 2d = 10 + 2 (5) = 10 + 10 = 20

t4

= a + 3d = 10 + 3 (5) = 10 + 15 = 25

One of the possible ways of writing the first four terms of A.P.could be 10, 15, 20, 25.

4. Write first three terms of A.P. whose common difference in 3.Sol. For an A.P. d = 3

Let us take a = 5t

1

= a = 5t

2= a + d = 5 + ( 3) = 5 3 = 2

t3

= a + 2d = 5 + 2 ( 3) = 5 6 = 1

One of the possible ways of writing first three terms of A.P. couldbe 5, 2, 1.

HOTS PROBLEMS(Problems for developing Higher Order Thinking Skill)

1. If the sum of p terms of an A. P. is equal to the sum of q terms thenshow that the sum of its p + q terms is zero. (5 marks)

Given : Sp = SqProve : S

p+q= 0

Proof : Sn

=n

2[2a + (n 1)d]

Sp

=p

2[2a + (p 1) d] ......(i)

Similarly,

Sq

=q

2[2a + (q 1) d] .....(ii)

Similarly,

Sp+q=

p q

2

+

[2a + (p + q 1) d] .....(iii) Sp = Sq [Given]


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SCHOOL SECTION44

p

2[2a + (p 1) d] =

q

2[2a + (q 1) d]

Multiplying both sides by 2,p [2a + (p 1) d] = q [2a + (q 1) d]

p [2a + pd d] = q [2a + qd d]

2ap + p2

d pd = 2aq + q 2

d qd 2ap + p2d pd 2aq q2d + qd = 0 2ap 2aq + p2d q2d pd + qd = 0 2a (p q) + d (p2 q2) d (p q) = 0 2a (p q) + d (p + q) (p q) d (p q) = 0

Dividing throughout by p q we get,2a + d (p + q) d = 0

2a + d [p + q 1] = 0 ......(iv)Substituting (iv) in (iii) we get,

Sp+q

=p q

2

+[0]

Sp+q

= 0

Hence proved.

2. How many two digit numbers leave the remainder 1 when divided by 5 ?(3 marks)

Sol. The two digit numbers that leave a remainder of 1 when divided by5 are as follows11, 16, 21, 26, 31, ............, 96.

These numbers form an A.P.With a = 11, d = 5Let, t

n= 96

tn

= a + (n 1) d 96 = 11 + (n 1) 5 96 = 11 + 5n 5 96 = 6 + 5n 96 6 = 5n 5n = 90

n =90

5 n = 18

There are 18 two digit numbers leaving a remainder of 1 when dividedby 5.

3. How many terms of the A. P. 16, 14, 12, ..... are needed to given thesum 60 ? Explain why we get double answer? (4 marks)

Sol. For the A.P. 16, 14, 12, .........a = 16d = 14 16 = 2Let, S

n= 60

Sn

=n

2[2a + (n 1) d]

60 =n

2[2 (16) + (n 1) ( 2)]

120 = n [32 2n + 2]

120 = n [34 2n]

20 = 34n 2n2

2n2

34n + 120 = 0Dividing throughout by 2,


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SCHOOL SECTION 45

n2 17n + 60 = 0

n2 12n 5n + 60 = 0

n (n 12) 5 (n 12) = 0

(n 12) (n 5) = 0

n 12 = 0 or n 5 = 0 n = 12 or n = 5 The no. of terms required to give a sum of 60 are 12 or 5.

The reason for getting double answers is the common difference is 2 and the progression is towards the negative side.The A.P. is as follows :16, 14, 12, 10, 8, 6, 4, 2, 0, 2, 4, 6 ....S

5= 16 + 14 + 12 + 10 + 8 = 60

S12

= 16 + 14 + 12 + 8 + 6 + 4 + 2 + 0 2 4 6 = 60

4. If the 9th term of an A. P. is zero then prove that the 29th term is doublethe 19th term. (4 marks)

Given : t9 = 0Prove : t

29= 2t

19

Proof : tn

= a + (n 1) d t

9= a + (9 1) d

0 = a + 8d a + 8d = 0 .......(i)

t29

= a + (29 1) d t

29= a + 28d

t29

= a + 8d + 20d t

29= 0 + 20d [From (i)]

t29

= 20d .......(ii)Similarly, t

19= a + (19 1)d

t19 = a + 18d t

19= a + 8d + 10d

t19

= 0 + 10d [From (i)] t

19= 10d ......(iii)

Dividing (ii) by (iii),

t

t29

19

=20d

10d

t

t29

19

= 2

t29

= 2t19

Hence proved.

5. If A and G are the A. M. and G. M. between two numbers then prove

that the numbers are A + +(A G) (A G) . (5 marks)

Sol. Let the two number be x and y.

A and G are A.M. and G.M. of x and y

A =x + y

2.... (i)

G = xy

G2 = xy

y =G

x

2

.... (ii)

Now, 2A = x + y


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SCHOOL SECTION46

2A x = y ....(iii)

G

x

2

= 2A x

G2 = 2Ax x 2

x2 2Ax G2 = 0

Comparing with ax2 + bx + c = 0 we get, a = 1, b = 2A, c = G2 b2 4ac = (2A)2 4 (1) G2

b2 4ac = 4A 2 4G2

= 4 (A 2 G2)= 4 (A + G) (A G)

x = b b 4ac

2a

2

x = ( 2A) 4 (A + G) (A G)

2 1

x =

2A 2 (A + G) (A G)

2

x =( )2 A (A G) (A G)

2

+

x = A + (A + G) (A G) ......(iv)

Substituting (iv) in (iii),

y = 2A [A (A + G) (A G) ]

y = 2A A (A + G) (A G)

y = A (A + G) (A G)

The two numbers are A (A + G) (A G)

6. Find the sum of all terms given in a sequence 1, 2 + x, 3x2,4 + x3, ...., n + xn1.(5 marks)

Sol. For the given sequence 1, 2 + x, 3 + x2, 4 + x3 ......, n + xn1

Sum of all terms of sequenceS = 1 + 2 + x + 3 + x2 + 4 + x3 + ...... + n + xn1

S = (1 + 2 + 3 + ..... + n) + (x + x2 + x3 + ...... + xn1)Consider 1 + 2 + 3 + ....... + n

These terms are in A.P. with a = 1, d = 1, tn

= n Sum of all terms of A.P. with n terms

=n

2 [t1 + tn]

=n

2[1 + n]

=n n

2

+2

......(i)

Consider, x + x2 + x3 + ..... + xn1

These terms form a G.P. with a = x, r = xSum of all terms of G.P. with n 1 terms

=( )a r 1r 1

n 1

= ( )x x 1x 1

n 1


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SCHOOL SECTION 47

Sum of all terms of given sequence= Sum of all terms of A.P. + sum of all terms of G.P.

=n n x (x 1)

2 x 1

++

2 n 1

7. If m times the mth term of an A. P. is equal to n times its nth term thenshow that the (m + n)th term of the A. P. is zero. (4 marks)

Given : mtm

= ntn

Prove : tm+ n

= 0Proof : t

n= a + (n 1) d .... (i)

Similarly tm

= a + (m 1) d .... (ii)Similarly t

m+ n= a + (m + n 1) d .... (iii)

m [a + (m 1) d] = n [a + (n 1) d] m [a + md d] = n [a + nd d] am + m2d md = an + n2d nd am an + m2d n2d md + nd = 0 a (m n) + d (m2 n2) d (m n) = 0 a (m n) + d (m + n) (m n) d (m n) = 0

Dividing through at by m n, we geta + d (m + n) = d = 0

a + d [m + n 1] = 0 a + (m + n 1) d = 0 t

m+n+ n = 0 [From (iii)]

Hence proved.

8. For a sequence Sn

= n(3n + 2), find tn

Examine whether the sequence isA. P or G. P. (4 marks)

Sol. Sn

= n (3n + 2) S

1= 1 (3 (1) + 2) = 1(5) = 5

S2

= 2 (3 (2) + 2) = 2 (8) = 16S

3= 3 (3 (3) + 2) = 3 (11) = 33

S4

= 4 (3 (4) + 2) = 4 (14) = 56t

1= S

1= 5

t2

= S2

S1

= 16 5 = 11t

3= S

3 S

2= 33 16 = 17

t4

= S4

S3

= 56 33 = 23Now,t2 t

1= 11 5 = 6

t3

t2

= 17 11 = 6t4

t3

= 23 17 = 6

The difference between two consecutive terms is 6 which is aconstant, the sequence is an A.P.with a = 5, d = 6We know for A.P. t

n= a + (n 1) d

tn

= 5 + (n 1) 6 t

n= 5 + 6n 6

tn

= 6n 1

9. For a sequence Sn

=

n n

n

4 3

3 , find tn Examine whether the sequence is

A. P or G. P. (5 marks)

Sol. Sn

=4 3

3

n n

n

S1

= 4 33

1 1

1 =4 3

3= 1

3


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S2

=4 3

3

2 2

2 =16 9

9=

7

9

S3

=4 3

3

3 3

3 =64 27

27=

37

27

t1

= S1

= 13

t2

= S2

S1

=7

9=

1

3

t2

=7

9

3

9

t2

=4

9

t3

= S3

S2

=37

27

7

9

t3 =37

27 21

27

t3

=16

27Now,

t

t2

1=

4

9

1

3

t

t2

1=

4

9

3

1

t

t2

1=

4

3

tt

3

2= 16

27 4

9

t

t3

2=

16

27

9

4

t

t3

2=

4

3

The ratio of two consecutive terms is4

3which is a constant, the

sequence is a G.P. with a =1

3

r = 43

We know for G.P. t

n= arn1

tn

=1

3

4

3

n1

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Ch. 1_Arithmetic Progression And