# Question-1

**Write the first 5 terms of each of the following sequences:**

(i) a

(ii) a

(iii) a

(iv) a

(v) a

(vi) a

(i) a

_{n}= (-1)^{n-1}5^{n+1}(ii) a

_{n }=(iii) a

_{n}= -11n +10(iv) a

_{n}=(v) a

_{n}=(vi) a

_{n}=**Solution:**

^{(i) }a

_{n}= (-1)

^{n-1}5

^{n+1}

a

_{1}= (-1)

^{0}5

^{2}= 5

^{2};

a

_{2}= (-1)

^{1}5

^{3 }= -5

^{3}

a

_{3}= (-1)

^{2}5

^{4}= 5

^{4};

a

_{4}= (-1)

^{3}5

^{5}= -5

^{5}

a

_{5}= (-1)

^{4}5

^{6}= 5

^{6 }

(ii)a_{n }=

a_{1} = ==

a_{2 }= ==

a_{3} = =

a_{4} = = 21

a_{5} = =

(iii)a_{n} = -11n +10

a_{1} = -11+ 10 = -1

a_{2} = -22 + 10 = -12

a_{3} = -33 + 10 = -23

a_{4} = -44 +10 = -34

a_{5} = -55 + 10 = -45

(iv) a_{n} **= = **

a_{2} = =

a_{3} = =

a_{4} = =

a_{5} = =

(v) a_{n} =

a_{1} = =

a_{2} = = 0

a_{3} = =

a_{4} = = 0

a_{5} = =

(vi) a_{n} =

a_{1} = ; a_{2} = ; a_{3} = ; a_{4} = ; a_{5} =

# Question-2

**Find the first terms of the following sequences whose n**

(ii) a

(iii) a

(iv) a

^{th }term is_{(i) }a_{n}= 2+; a_{5, }a_{7}(ii) a

_{n}= cos; a_{4, }a_{5}(iii) a

_{n}= ; a_{7, }a_{10}(iv) a

_{n}= (-1)^{n-1}2^{n+1}; a_{5, }a_{8 }**Solution:**

(i) a

_{n}= 2 +

a

_{5}= 2 +

= ;

a

_{7}= 2 +

=

(ii) a

_{n}= cos;

a

_{4 }=

_{ }cos

_{ }= cos 2Ï€ = 1

a

_{5 }= cos= cos

_{ }= cos

_{ }=

_{ }0

(iii) a

_{n}=

a

_{7}=

= ;

a

_{10}=

=

(iv) a_{n} = (-1)^{n-1} 2^{n+1
}a_{5 }= (-1)^{4} 2^{5+1}

= 2^{6}

= 64;

a_{8} = (-1)^{7} 2^{8+1}

= - 2^{9}

= -512

# Question-3

**Find the first 6 terms of the sequence whose general term is**

a

a

_{n}= {n^{2}-1 if n is odd**if n is even}**

**Solution:**

a

_{1}= 1

^{2}â€“ 1 = 0

a

_{2}= =

a

_{3 }= 3

^{2}â€“ 1 = 8

a

_{4}= =

a

_{5}= 5

^{2}â€“ 1 = 24

a

_{6}= =

# Question-4

**Write the first five terms of the sequence given by**

(i) a

(ii) a

(iii) a

(iv) a

(i) a

_{1}= a_{2}= 2, a_{n}= a_{n-1}â€“1, n>2(ii) a

_{1}= 1, a_{2}= 2, a_{n}= a_{n-1}+ a_{n-2}, n>2(iii) a

_{1}= 1, a_{n}= na_{n-1}, nâ‰¥ 2(iv) a

_{1}= a_{2 }= 1, a_{n}= 2a_{n-1}+ 3a_{n-2,}n>2**Solution:**

(i) Put n = 3 a

_{3}= a

_{2}â€“1 = 2-1 = 1

n = 4 a

_{4 }= a

_{3}â€“1 = 1-1 = 0

n = 5 a

_{5}= a

_{4}â€“1 = 0-1 = -1

(ii) Put n = 3a_{3} = a_{2} + a_{1} = 2 + 1 = 3

n = 4a_{4} = a_{3} + a_{2} = 3 + 2 = 5

n = 5a_{5} = a_{4} + a_{3} = 5 + 3 = 8

(iii) Put n = 2a_{2} = 2 a_{1} = 2.1 = 2

n = 3a_{3} = 3 a_{2} = 3.2 = 6

n = 4a_{4} = 4 a_{3} = 4.6 = 24

n = 5a_{5} = 5 a_{4} = 5.24 = 120

(iv) ** **Put n = 3 a_{3} = 2a_{2} + 3a_{1} = 2(1) + 3(1) = 5

n = 4 a_{4} = 2a_{3} + 3a_{2} = 2(5) + 3(1) = 13

n = 5 a_{5} = 2a_{4} + 3a_{3} = 2(13) + 3(5) = 41

# Question-5

**Find the n**

^{th}partial sum of the series**Solution:**

S

_{n}= + + + â€¦â€¦â€¦â€¦.

S

_{n+1}= ++ â€¦â€¦â€¦â€¦.. + +

S

_{n+1}= s

_{n}+

S

_{n+1}= + + + â€¦â€¦â€¦â€¦. +

= =

s

_{n}+ = +s

_{n}

3 s

_{n}+ = 1 + s

_{n }2s

_{n}= 1-

s

_{n}=

# Question-6

**Find the sum of first n terms of the series 5**

^{n }**Solution:**

5

^{n}= 5 + 5

^{2}+ 5

^{3}+â€¦â€¦â€¦..+ 5

^{n}+â€¦â€¦â€¦

s

_{n}= 5 + 5

^{2}+ 5

^{3}+â€¦â€¦â€¦..+ 5

^{n}

s

_{n+1}= 5 + 5

^{2}+ 5

^{3}+â€¦â€¦â€¦..+ 5

^{n}+ 5

^{n+1 }= s

_{n}+ 5

^{n+1 }

Also s

_{n+1}= 5 + 5

^{2}+ 5

^{3}+â€¦â€¦â€¦..+ 5

^{n}+ 5

^{n+1 }= 5[1 + 5 + 5

^{2}+â€¦â€¦â€¦â€¦..+ 5

^{n}]

= 5[1 + s

_{n}]

s

_{n}+ 5

^{n+1}= 5 + 5 s

_{n }4s

_{n}= 5

^{n+1}-5

s

_{n}=

# Question-7

**Find the sum of 101**

^{th}term to 200^{th}term of the series**Solution:**

To find S

_{200}â€“ S

_{100 }

To find S

_{200: }S

_{200}= ++ +â€¦â€¦â€¦.+

S

_{201}= ++ +â€¦â€¦â€¦.++

= s

_{200}+

also, S

_{201}= ++ +â€¦â€¦â€¦.++

=

S

_{200}+ =

2S

_{200}+ = 1 + S

_{200 }S

_{200}= 1-

Similarly S

_{100}= 1-

Hence S

_{200}â€“ S

_{100}= -

= -

# Question-8

**Find five arithmetic means between 1 and 19.**

**Solution:**

Let 1, x

_{1}, x

_{2},

_{ }x

_{3}, x

_{4}, x

_{5}, 19 be in A.P.

Let d be the common difference

19 = 1 + (n-1)d

19 = 1 + 6d

d = 3

x

_{1}= 1 + 3 = 4

x

_{2}= 4 + 3 = 7

x

_{3}= 7 + 3 = 10

x

_{4}= 10 + 3 = 13

x

_{5}= 13 + 3 = 16

The arithmetic means are 4, 7, 10, 13, 16.

# Question-9

**Find six arithmetic mean between 3 and 17.**

**Solution:**

Let 3, x

_{1}, x

_{2},

_{ }x

_{3}, x

_{4}, x

_{5}, x

_{6}, 17 be in A.P

Then 17 = 3 + (n-1)d

17 = 3 + 7d

14 = 7d

d = 2

x

_{1}= 3 + 2 = 5

x

_{2 }= 5 + 2 = 7

x

_{3}= 7 + 2 = 9

x

_{4}= 9 + 2 = 11

x

_{5}= 11 + 2 = 13

x

_{6}= 13 + 2 = 15

The arithmetic means are 5, 7, 9, 11, 13, 15.

# Question-10

**Find the single A.M. between**

(i) 7 and 13

(ii) 5 and â€“3

(iii) (p + q) and (p - q)

(i) 7 and 13

(ii) 5 and â€“3

(iii) (p + q) and (p - q)

**Solution:**

(i)

**A.M. between 7 and 13 = = 10**

(ii)

**A.M. between 5 and â€“3 = = 1**

(iii)

**A.M. between (p + q) and (p â€“ q) = = p**

# Question-11

**If b is the G.M. of a and c and x is the A.M of a and b and y is the A.M of b and c, prove that + = 2.**

**Solution:**

b = G.M. of a and c = b â€¦â€¦â€¦â€¦..(1)

x = A.M. of a and b x = â€¦â€¦â€¦â€¦â€¦(2)

y = A.M. between b and c y = â€¦â€¦..(3)

To prove that += 2

From (1) b

^{2}= ac c =

+ = +

= +

= +

= +

=

= 2

# Question-12

**The first and second terms of H.P are and respectively, find the 9**

^{th}term.**Solution:**

Let the H.P are ,, +â€¦â€¦â€¦..

= a = 3

= 5 = a + d; d = 2

9

^{th}term = = =

# Question-13

**If a, b, c are in H.P., prove that + = 2.**

**Solution:**

If a, b, c are in H.P then b =

=

**= =**â€¦â€¦â€¦â€¦â€¦â€¦.(1)

Also

**=**

= =â€¦â€¦â€¦â€¦â€¦â€¦.(2)

= =

Adding (1) and (2)

**+ = +**

**= -**

=

=

= 2

=

=

Hence + =2.

# Question-14

**The difference between two positive numbers is 18, and 4 times their G.M. is equal to 5 times their H.M. find the numbers.**

**Solution:**

Let the two numbers be a and b

b - a = 18

4 = 5

2 =

2(a + b) = 5

4(a + b)

^{2}= 25ab

4(a

^{2 }+ b

^{2 }+ 2ab) = 25ab

4a

^{2}+ 4b

^{2}â€“17ab = 0

4a

^{2}+ 4(18 + a)

^{2}â€“ 17a(18 + a) = 0

4a

^{2}+ 4(324 + 36a + a

^{2})

^{ }- 306a - 17a

^{2}= 0

4a

^{2}+ 1296 + 144a + 4a

^{2}â€“ 306a â€“ 17a

^{2}= 0

-9a

^{2 }-162a + 1296 = 0

a

^{2}+ 18a -144 = 0

(a + 24)(a - 6) = 0

a = -24 (or) 6

If a = 6 then b is 24.

Therefore the numbers are 6 and 24.

# Question-15

**If the A.M. between two numbers is 1, prove that their H.M. is the square of their G.M.**

**Solution:**

A.M. between two numbers a and b is 1.

= 1

a + b = 2

HM = (GM)

^{2 }

HM = = = ab

GM =

(GM)

^{2}= ab

Hence HM = GM

^{2}

# Question-16

**If a, b, c are in A.P and a, mb, c are in G.P then prove that a, m**

^{2}b, c are in H.P.**Solution:**

**Given**

a, b, c are in A.P.

b = â€¦â€¦â€¦(1)

a, mb, c are in G.P.

mb = â€¦â€¦â€¦â€¦.(2)

**To prove**

a,m

^{2}b,c are in H.P.

i.e., m

^{2}b =

**Proof**

R.H.S = = from (2) and (1)

= m

^{2}b = LHS

# Question-17

**If the p**

^{th}and q^{th}terms of a H.P. are q and p respectively, show that (pq)^{th}term is 1.**Solution:**

**Given**

p

^{th}and q

^{th}terms of a H.P. are q and p.

Therefore = q â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)

and = p â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(2)

**To prove**

pq

^{th}term, ie., = 1

**Proof**

From (1) a + pd â€“ d =

From (2) a + qd â€“ d =

Subtracting (p - q)d = - =

d =

a + p - =

a =

a + (pq â€“ 1)d = + (pq - 1)

=

= = 1

= 1

pq^{th} term is 1.

# Question-18

**Three number form a H.P. the sum of the numbers is 11 and the sum of the reciprocals is one. Find the numbers.**

**Solution:**

Let ,,be in H.P.

Their sum is + + = 11

The sum of their reciprocal is a - d + a + a + d = 1

3a = 1

a =

**+**+ = 11

+ 3 + = 11

+ = 8

= 8

= 8

6 = 8 - 72d

^{2 }72d

^{2}= 2

d

^{2}=

d =

The numbers are + + =

The numbers are 6, 3, 2.

# Question-19

**Write the first four terms in the expansions of the following:**

(i) where |x| >2

(ii) where |x|<2

(i) where |x| >2

(ii) where |x|<2

**Solution:**

(i)

**= =**

**= **

**= **

(ii)**= = **

**= **

**= **

# Question-20

**Evaluate the following:**

(i) correct to 4 places of decimals.

(ii) correct to 4 places of decimals.

(iii) - correct to 3 place of decimals.

(i) correct to 4 places of decimals.

(ii) correct to 4 places of decimals.

(iii) - correct to 3 place of decimals.

**Solution:**

(i) = (1003)

^{ }= (1000 + 3)

^{}

= (1000)

^{1/3 }= 10

^{ }= 10

^{ }= 10

^{ }= 10.00999

(ii) = = = =

= (1 + 0.024) =

**=**= 0.1984256

(iii)

**-**= (1000 + 3)

^{1/3}- (1000 â€“ 3)

^{ 1/3 }= 10 - 10

= 10 [1 + 0.003] - 10[1 - 0.003]

= 10

-10

= 10

= 10[0.002] = 0.02

# Question-21

**If x is so small show that**

(i)

(ii) =1 - 4x(app.)

(i)

(ii) =1 - 4x(app.)

**Solution:**

(i) = (1-x)

^{ }(1+x)

=

=

= 1 -

= 1- x + (app.)

(ii)** **= (1 + x)^{-2}(1 + 4x)^{-1/2 }

=

= (1 - 2x + â€¦â€¦)(1 â€“ 2x + â€¦â€¦.)

= 1 - 2x - 2x + 4x^{2 }+â€¦. = 1 - 4x(app.)

# Question-22

**If x is so large prove that - = nearly.**

**Solution:**

**- =**x- x

**= x- x**

= x +

=

**= =**approximately

# Question-23

**If c is small compared to l, show that + = 2 +(app)**

**Solution:**

**+ = +**

= +

= +

Since c is small in comparison with l then || < 1,âˆ´ binomial expansion is valid.

= 1++

+1++

= 1 -

= 2 +approximately.

# Question-24

**Find the 5**

^{th }term in the expansion of (1 - 2x^{3})^{11/2}.**Solution:**

(1 - 2x

^{3})

^{11/2 }= 1 +

5

^{th}term is = x

^{12}

# Question-25

**Find the (r + 1)**

^{th}term in the expansion of (1 - x)^{-4}.**Solution:**

T

_{r+1 }in (1 - x)

^{-4 }

(1-x)

^{-4}= [1.2.3 + 2.3.4.x +â€¦â€¦â€¦â€¦..(r + 1)(r + 2)(r + 3)x

^{r}+â€¦â€¦â€¦]

T

_{r+1 }=

# Question-26

**Show that x**

^{n}= 1 + n**Solution:**

R.H.S = 1 + n

Put y = 1-

= 1 + ny+

= (1 - y)

^{-n }=

=

= x

^{n }= L.H.S

# Question-27

**Find the sum to infinity of the series**

(i) 1+

(ii) 1 -

(iii)

(i) 1+

(ii) 1 -

(iii)

**Solution:**

(i) Let S = 1+

= 1++â€¦â€¦..

=

= = 4= 4

^{1}4

^{}= = 4(2) = 8

(ii) Let S = 1 -

= 1 -

**=**

=

=

=

=

(iii) Let S =

S + 1 = 1+

= 1+

= == 2^{3/2
}S + 1 = 2^{3/2}

Therefore S = 2^{3/2} â€“ 1

# Question-28

**Show that the coefficient of x**

^{n}in the infinite series**1 + is .**

(ii) Show that = 1 + .

(iii) Show that 2 = n + c.

(ii) Show that = 1 + .

(iii) Show that 2 = n + c.

**Solution:**

(i)

**1 + = e**

^{y}= e

^{b + ax }= e

^{b}. e

^{ax }= e

^{b}

Coefficient of x

^{n}= e

^{b}.

(ii) L.H.S = =

=

=

=

= R.H.S

(iii) L.H.S = 2

Put log n = y

2 = 2

= e^{y} + e^{-y}

= e^{logn} + e^{-logn}

= e^{ logn + }e ^{log1/n}

= n +

# Question-29

**Show that log a â€“ log b = + ++ â€¦â€¦â€¦.**

**Solution:**

R.H.S = + ++ â€¦â€¦â€¦.

Put y =

y + ++â€¦â€¦â€¦.. = - log (1 â€“ y)

= - log

= - log

= log

= log a â€“ log b

= L.H.S

# Question-30

**Prove that log =**

**Solution:**

R.H.S =

Put = y

y + +â€¦â€¦â€¦.. = log

= log

= log

= log

= log

= L.H.S

# Question-31

**Find the sum to infinity the series +++â€¦â€¦â€¦â€¦â€¦.**

**Solution:**

+++â€¦â€¦â€¦â€¦â€¦.

Put y =

+++â€¦â€¦â€¦â€¦ = log

= log

= log

= log

= log

# Question-32

**If x is so small show that**

(i)

(ii) =1 - 4x(app.)

(i)

(ii) =1 - 4x(app.)

**Solution:**

(i)

**= (1-x)**

^{ }(1+x)

=

=

= 1 -

= 1- x + (app.)

(ii)** ** = (1 + x)^{-2}(1 + 4x)^{-1/2 }

=

= (1 - 2x + â€¦â€¦)(1 â€“ 2x + â€¦â€¦.)

= 1 - 2x - 2x + 4x^{2 }+â€¦. = 1 - 4x(app.)

# Question-33

**If x is so large prove that - = nearly.**

**Solution:**

**- =**x- x

**= x- x**

= x +

=

**= =**approximately

# Question-34

**If c is small compared to l, show that + = 2 +(app)**

**Solution:**

**+ = +**

= +

= +

Since c is small in comparison with l then || < 1,âˆ´ binomial expansion is valid.

= 1++

+1++

= 1 -

= 2 +approximately.

# Question-35

**Find the 5**

^{th }term in the expansion of (1 - 2x^{3})^{11/2}.**Solution:**

(1 - 2x

^{3})

^{11/2 }= 1 +

5

^{th}term is = x

^{12}

# Question-36

**Find the (r + 1)**

^{th}term in the expansion of (1 - x)^{-4}.**Solution:**

T

_{r+1 }in (1 - x)

^{-4 }

(1-x)

^{-4}= [1.2.3 + 2.3.4.x +â€¦â€¦â€¦â€¦..(r + 1)(r + 2)(r + 3)x

^{r}+â€¦â€¦â€¦]

T

_{r+1 }=

# Question-37

**Show that x**

^{n}= 1 + n**Solution:**

R.H.S = 1 + n

Put y = 1-

= 1 + ny+

= (1 - y)

^{-n }

=

=

= x^{n}

= L.H.S

# Question-38

**Find the sum to infinity of the series**

(i) 1+

(ii) 1 -

(iii)

(i) 1+

(ii) 1 -

(iii)

**Solution:**

(i) Let S = 1+

= 1++â€¦â€¦..

=

= = 4= 4

^{1}4

^{}= = 4(2) = 8

(ii)Let S = 1 -

= 1 -

** =
=
= **

(iii) Let S =

S + 1 = 1+

= 1+

= == 2^{3/2
}S + 1 = 2^{3/2}

Therefore S = 2^{3/2} â€“ 1

# Question-39

**(i) Show that the coefficient of x**

^{n}in the infinite series**1 + is .**

(ii) Show that = 1 + .

(iii) Show that 2 = n + c.

(ii) Show that = 1 + .

(iii) Show that 2 = n + c.

**Solution:**

(i)

**1 + = e**

^{y}= e

^{b + ax }= e

^{b}. e

^{ax }= e

^{b}

Coefficient of x

^{n}= e

^{b}.

(ii) L.H.S = =

=

=

=

= R.H.S

(iii) L.H.S = 2

Put log n = y

2 = 2

= e^{y} + e^{-y}

= e^{logn} + e^{-logn}

= e^{ logn + }e ^{log1/n}

= n +

# Question-40

**Show that log a â€“ log b = + ++ â€¦â€¦â€¦.**

**Solution:**

R.H.S = + ++ â€¦â€¦â€¦.

Put y =

y + ++â€¦â€¦â€¦.. = - log (1 â€“ y)

= - log

= - log

= log

= log a â€“ log b

= L.H.S

# Question-41

**Prove that log =**

**Solution:**

R.H.S =

Put = y

y + +â€¦â€¦â€¦.. = log

= log

= log

= log

= log

= L.H.S

# Question-42

**Find the sum to infinity the series +++â€¦â€¦â€¦â€¦â€¦.**

**Solution:**

+++â€¦â€¦â€¦â€¦â€¦.

Put y =

+++â€¦â€¦â€¦â€¦ = log

= log

= log

= log

= log