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Question-1

Write the first 5 terms of each of the following sequences:
 (i) an = (-1)n-1 5n+1
(ii) an =
(iii) an = -11n +10

(iv) an =

(v) an =
(vi) an =

Solution:
(i) an = (-1)n-1 5n+1
a1 = (-1)052 = 52;
a2 = (-1)1 53 = -53
a3 = (-1)254 = 54;
a4 = (-1)3 55 = -55
a5 = (-1)456 = 56

(ii)an =
a1 = ==
a2 = ==
a3 = =
a4 = = 21
a5 = =

(iii)an = -11n +10
a1 = -11+ 10 = -1
a2 = -22 + 10 = -12
a3 = -33 + 10 = -23
a4 = -44 +10 = -34
a5 = -55 + 10 = -45

(iv) an = =
a2 = =
a3 = =
a4 = =
a5 = =

(v) an =
a1 = =
a2 = = 0
a3 = =
a4 = = 0
a5 = =

(vi) an =
a1 = ; a2 = ; a3 = ; a4 = ; a5 =

Question-2

Find the first terms of the following sequences whose nth term is
(i) an = 2+; a5, a7
(ii) an = cos; a4, a5
(iii) an = ; a7, a10
(iv) an = (-1)n-1 2n+1; a5, a8

Solution:
(i) an = 2 +
        a5 = 2 +
            = ;
        a7 = 2 +
            =

(ii) an = cos;
      a4 = cos = cos 2π = 1
      a5 = cos= cos = cos = 0

(iii) an =
      a7 =
          = ;
     a10 =
          =

(iv) an = (-1)n-1 2n+1
      
a5 = (-1)4 25+1
          = 26
          = 64;
      a8 = (-1)7 28+1
          = - 29
          = -512

Question-3

Find the first 6 terms of the sequence whose general term is
an = {n2-1 if n is odd
if n is even}

Solution:
a1 = 12 – 1 = 0
a2 = =
a3 = 32 – 1 = 8
a4 = =
a5 = 52 – 1 = 24
a6 = =

Question-4

Write the first five terms of the sequence given by
(i) a1 = a2 = 2, an = an-1 –1, n>2
(ii) a1 = 1, a2 = 2, an = an-1 + an-2, n>2
(iii) a1 = 1, an = nan-1, n 2
(iv) a1 = a2 = 1, an = 2an-1 + 3an-2, n>2

Solution:
(i) Put n = 3 a3 = a2 –1 = 2-1 = 1
             n = 4 a4 = a3 –1 = 1-1 = 0
             n = 5 a5 = a4 –1 = 0-1 = -1

(ii) Put n = 3a3 = a2 + a1 = 2 + 1 = 3
           n = 4a4 = a3 + a2 = 3 + 2 = 5
           n = 5a5 = a4 + a3 = 5 + 3 = 8

(iii) Put n = 2a2 = 2 a1 = 2.1 = 2
           n = 3a3 = 3 a2 = 3.2 = 6
           n = 4a4 = 4 a3 = 4.6 = 24
           n = 5a5 = 5 a4 = 5.24 = 120

(iv)   Put n = 3 a3 = 2a2 + 3a1 = 2(1) + 3(1) = 5
             n = 4 a4 = 2a3 + 3a2 = 2(5) + 3(1) = 13
             n = 5 a5 = 2a4 + 3a3 = 2(13) + 3(5) = 41

Question-5

Find the nth partial sum of the series

Solution:

   Sn = + + + ………….
Sn+1 = ++ ………….. + +
Sn+1 = sn +
Sn+1 = + + + …………. +
       = =
sn + = +sn
3 sn + = 1 + sn
         
2sn = 1-
          sn =

Question-6

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

Solution:
5n = 5 + 52 + 53 +………..+ 5n +………
        sn = 5 + 52 + 53 +………..+ 5n
     sn+1 = 5 + 52 + 53 +………..+ 5n + 5n+1
             
= sn + 5n+1
Also sn+1 = 5 + 52 + 53 +………..+ 5n + 5n+1
                
= 5[1 + 5 + 52 +…………..+ 5n ]
              = 5[1 + sn]
sn + 5n+1 = 5 + 5 sn
         
4sn = 5n+1-5
       sn =

Question-7

Find the sum of 101th term to 200th term of the series

Solution:

To find S200 – S100
To find S200: S200 = ++ +……….+
                  S201 = ++ +……….++
                         = s200 +
           also, S201 = ++ +……….++
                         =
         S200 + =
       2S200 + = 1 + S200
                     
S200 = 1-
      Similarly S100 = 1-
Hence S200 – S100 = -
                         = -

Question-8

Find five arithmetic means between 1 and 19.

Solution:
Let 1, x1, x2, x3, x4, x5, 19 be in A.P.
Let d be the common difference
19 = 1 + (n-1)d
19 = 1 + 6d
d = 3
x1 = 1 + 3 = 4
x2 = 4 + 3 = 7
x3 = 7 + 3 = 10
x4 = 10 + 3 = 13
x5 = 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, x1, x2, x3, x4, x5, x6, 17 be in A.P
Then 17 = 3 + (n-1)d
17 = 3 + 7d
14 = 7d
 d = 2
x1 = 3 + 2 = 5
x2 = 5 + 2 = 7
x3 = 7 + 2 = 9
x4 = 9 + 2 = 11
x5 = 11 + 2 = 13
x6 = 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 –3
(iii) (p + q) and (p - q)

Solution:
(i) A.M. between 7 and 13 = = 10
(ii) A.M. between 5 and –3 = = 1
(iii) A.M. between (p + q) and (p – 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 …………..(1)
x = A.M. of a and b x = ……………(2)
y = A.M. between b and c y = ……..(3)
To prove that += 2
From (1) b2 = ac c =
         + = +
                 = +
                 = +
                 = +
                 = 
                 = 2

Question-12

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

Solution:
Let the H.P are ,, +………..
         = a = 3
      = 5 = a + d; d = 2
9th 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 =
=
= = ……………….(1)
Also =
= =
……………….(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(a2 + b2 + 2ab) = 25ab
4a2 + 4b2 –17ab = 0
4a2 + 4(18 + a)2 – 17a(18 + a) = 0
4a2 + 4(324 + 36a + a2) - 306a - 17a2 = 0
4a2 + 1296 + 144a + 4a2 – 306a – 17a2 = 0
-9a2 -162a + 1296 = 0
a2 + 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 = GM2

Question-16

If a, b, c are in A.P and a, mb, c are in G.P then prove that a, m2b, c are in H.P.

Solution:
Given
a, b, c are in A.P.
  b = ………(1)
a, mb, c are in G.P.
mb = ………….(2)

To prove
a,m2b,c are in H.P.
i.e., m2b =
Proof
 
R.H.S = = from (2) and (1)
         = m2 b = LHS

Question-17

If the pth and qth terms of a H.P. are q and p respectively, show that (pq)th term is 1.

Solution:
Given
pth and qth terms of a H.P. are q and p.
Therefore = q ………………….(1)
         and = p ……………………(2)

To prove
pqth term, ie., = 1

Proof
From (1) a + pd – d =
From (2) a + qd – d =
Subtracting (p - q)d = - =
                      d =
         a + p - =
                         a =
      a + (pq – 1)d = + (pq - 1)
                            =
                            = = 1
              = 1
pqth 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 - 72d2
                     
72d2 = 2
                   d2 =
                    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

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.

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 – 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.)

Solution:
(i) = (1-x) (1+x)
=
=
= 1 -
= 1- x + (app.)

(ii) = (1 + x)-2(1 + 4x)-1/2
=
= (1 - 2x + ……)(1 – 2x + …….)
= 1 - 2x - 2x + 4x2 +…. = 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, binomial expansion is valid.
= 1++
+1++
= 1 -
= 2 +approximately.

Question-24

Find the 5th term in the expansion of (1 - 2x3)11/2.

Solution:
(1 - 2x3)11/2 = 1 +
5th term is = x12

Question-25

Find the (r + 1)th term in the expansion of (1 - x)-4.

Solution:
Tr+1 in (1 - x)-4
(1-x)-4 = [1.2.3 + 2.3.4.x +…………..(r + 1)(r + 2)(r + 3)xr+………]
  Tr+1 =

Question-26

Show that xn= 1 + n

Solution:
R.H.S = 1 + n
Put y = 1-
        = 1 + ny+
        = (1 - y)-n
         
=
        =
        = xn
         
= L.H.S

Question-27

Find the sum to infinity of the series
(i) 1+
(ii) 1 -
(iii)

Solution:
(i) Let S = 1+
               = 1++……..
               =
     = = 4= 41 4= = 4(2) = 8

(ii) Let S = 1 -
             = 1 -
                =
                =
               =

(iii) Let S =
     S + 1 = 1+
              = 1+
              = == 23/2
      
S + 1 = 23/2

Therefore S = 23/2 – 1

Question-28

Show that the coefficient of xn in the infinite series 1 + is .
(ii) Show that = 1 + .
(iii) Show that 2 = n + c.

Solution:
(i) 1 + = ey = eb + ax = eb . eax = eb
           Coefficient of xn = eb .

(ii) L.H.S = =
             =
             =
             =
             = R.H.S

(iii) L.H.S = 2
 Put log n = y

2 = 2
                           = ey + e-y
                           = elogn + e-logn
                           = e logn + e log1/n
                           = n +

Question-29

Show that log a – log b = + ++ ……….

Solution:
R.H.S = + ++ ……….
    Put y =
y + ++……….. = - log (1 – y)
                           = - log
                           = - log
                           = log
                           = log a – log b
                           = L.H.S

Question-30

Prove that log =

Solution:
R.H.S =
Put = y
y + +……….. = log
                         = log
                         = log
                         = log
                         = log
                         = L.H.S

Question-31

Find the sum to infinity the series +++…………….

Solution:
+++…………….
Put y =
+++………… = log
                              = log
                              = log
                              = log
                              = log

Question-32

If x is so small show that

(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 + ……)(1 – 2x + …….)
                    = 1 - 2x - 2x + 4x2 +…. = 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, binomial expansion is valid.
                    = 1++
                   +1++
                   = 1 -
                   = 2 +approximately.

Question-35

Find the 5th term in the expansion of (1 - 2x3)11/2.

Solution:
(1 - 2x3)11/2 = 1 +
5th term is = x12

Question-36

Find the (r + 1)th term in the expansion of (1 - x)-4.

Solution:
Tr+1 in (1 - x)-4
(1-x)-4 = [1.2.3 + 2.3.4.x +…………..(r + 1)(r + 2)(r + 3)xr+………]
  Tr+1 =

Question-37

Show that xn= 1 + n

Solution:
R.H.S = 1 + n
Put y = 1-
        = 1 + ny+
        = (1 - y)-n

        =
        =
        = xn

        = L.H.S

Question-38

Find the sum to infinity of the series
(i) 1+
(ii) 1 -
(iii)

Solution:
(i) Let S = 1+
               = 1++……..
               =
     = = 4= 41 4= = 4(2) = 8

(ii)Let S = 1 -
            = 1 -
              =
              =
              =

(iii) Let S =
     S + 1 = 1+
              = 1+
              = == 23/2
      
S + 1 = 23/2
Therefore S = 23/2 – 1

Question-39

(i) Show that the coefficient of xn in the infinite series 1 + is .
(ii) Show that = 1 + .
(iii) Show that 2 = n + c.

Solution:
(i) 1 + = ey = eb + ax = eb . eax = eb
Coefficient of xn = eb .
(ii) L.H.S = =
                                   =
                                   =
                                   =
                                   = R.H.S

(iii) L.H.S = 2
Put log n = y
2 = 2
                           = ey + e-y
                           = elogn + e-logn
                           = e logn + e log1/n
                           = n +

Question-40

Show that log a – log b = + ++ ……….

Solution:
R.H.S = + ++ ……….
    Put y =
y + ++……….. = - log (1 – y)
                           = - log
                           = - log
                           = log
                           = log a – log b
                           = L.H.S

Question-41

Prove that log =

Solution:
R.H.S =
Put = y

y + +……….. = log
                        = log
                        = log
                        = log
                        = log
                        = L.H.S

Question-42

Find the sum to infinity the series +++…………….

Solution:
+++…………….
Put y =
+++………… = log
                              = log
                              = log
                              = log
                              = log




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