Skip to main content

WORKSHEET 1 Arithmetic Progressions

1.  In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
  • Solution:(i) The taxi fare after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
    The taxi fare will be 15, 15 + 8, 15 + 16, 15 + 24, 15 + 32, 15 + 40 etc.
    (i.e.,) 15, 23, 31, 39, 47, 55 etc.
    This is an AP as every succeeding term is obtained by adding 8 in its preceding term.
    (ii) The amount of air present in a cylinder when a vacuum pump removes  of the air remaining in the cylinder at a time.
    Let amount of air present in the cylinder be "a".
    The vacuum removes  of air remaining in the cylinder.
    \ The air present in the cylinder will be a, a - = , -== ,
    a, , , …
    This is not an AP because the difference between second term and first term is not the same as the difference between the third term and the second term.
    (iii) The cost of digging a well after every metre of digging, when it costs Rs. 150 for the first metre and rises by Rs. 50 for each subsequent metre.
    The cost of digging the well will be Rs. 150, Rs. (150 + 50), Rs. (150 + 100), Rs. (150 + 150) etc.
    (i.e.,) Rs. 150, Rs. 200, Rs. 250, Rs. 300, Rs. 350 etc
    This is an AP as every succeeding term is obtained by adding Rs. 50 in its preceding term.
    (iv) The amount of money in the account every year, when Rs. 10000 is deposited at compound interest at 8% per annum.
    Initial amount of money = Rs. 10,000
    Compound Interest = 8% p.a
    Amount every year increase by 8% Compound Interest p.a.
    In the second year amount increases to 10,000
    In the 3rd year amount increases to 10,000 
    In the 4th year amount increases to 10,000
    This is not an AP as the difference is not the same in this problem.
 2.  Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10 (ii) a = -2, d = 0 (iii) a = 4, d = -3 (iv) a = -1, d = 
  • Solution:(i) a = 10, d = 10
    Let a1, a2, a3, a4 be the first four terms of the AP
    Here a = a1 = 10
    a2 = a1 + d = 10 + 10 = 20
    a3 = a2 + d = 20 + 10 = 30
    a4 = a3 + d = 30 + 10 = 40
    \ The first four terms of the AP are 10, 20, 30, 40.
    (ii) a = -2, d = 0
    Let a1, a2, a3, a4 be the first four terms of the AP
    Here a = a1 = -2
    a2 = a1 + d = -2 + 0 = -2
    a3 = a2 + d = -2 + 0 = -2
    a4 = a3 + d = -2 + 0 = -2
    Thus first four terms of the AP are -2, -2, -2, -2
    (iii) a = 4, d = -3
    Let a1, a2, a3, a4 be the first four terms of the AP
    Here a = a1 = 4
    a2 = a1 + d = 4 + (-3) = 1
    a3 = a2 + d = 1 + (-3) = -2
    a4 = a3 + d = -2 + (-3) = -5
    The first 4 terms of the AP are 4, 1, -2, -5
    (iv) a = -1 , d =
    Let a1, a2, a3, a4 be the first four terms of the AP
    Here a = a1 = -1
    a2 = a1 + d = -1 +  = - 
    a3 = a2 + d = - +  = 0
    a4 = a3 + d = 0 +  = 
    The first 4 terms of the AP are –1, -, 0, .
    (v) a = -1.25, d = -0.25
    Let a1, a2, a3, a4 be the first four terms of the AP
    Here a = a1 = -1.25
    a2 = a1 + d = -1.25 + (-0.25) = -1.50
    a3 = a2 + d = -1.50 + (-0.25) = -1.75
    a4 = a3 + d = -1.75 + (-0.25) = -2.0
    The first 4 terms of the AP are -1.25, -1.50, -1.75, -2.0.
 3.  For the following APs, write the first term and the common difference:
  • Solution:(i) The first term is 3 and common difference = a2 – a1= 1 - 3 = -2.
    (ii) The first term is -5 and common difference = a2 – a1= (-1) – (-5) = 4.
    (iii) The first term is  and common difference = a2 – a1=  -  = .
    (iv) The first term is 0.6 and common difference = a2 – a1= (1.7) – (0.6) = 1.1.
 4.  Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, … (ii) 2, , … (iii) –1.2, -3.2, -5.2, -7.2, …
  • Solution:(i) 2, 4, 8, 16, …
    Here a2 – a1 = 4 – 2 = 2
    a3 – a2 = 8 – 4 = 4
    a4 – a3 = 16 – 8 = 8
    Since ak+1 - ak is not the same throughout, it is not an AP.
    (ii) 2, , …
    Here
    a2 – a1 =  – 2
    = 
    a3 – a2 = 3 –  = 
    a4 – a3 =  – 3 = 
    As ak+1 - ak is the same throughout, it is an AP with first term = 2 and common difference = 
    The next three terms are 4, 4 +  = ,  +  = = 5.
    (iii) –1.2, -3.2, -5.2, -7.2, …
    Here
    a2 – a1 = -3.2 – (-1.2)
    = -3.2 + 1.2 = -2
    a3 – a2 = -5.2 – (-3.2)
    = -2

    a4 – a3 = -7.2 – (-5.2)
    = -2
    Since ak+1 – ak is the same throughout it is an AP with first term (a) = -1. 2 and common difference, (d) = -2
    Thus the next three terms are
    a5 = a4 + d = –7.2 – 2 = -9.2
    a6 = a5 + d = -9.2 – 2 = -11.2
    a7 = a6 + d = - 11.2 – 2 = - 13.2
    (iv) –10, -6, -2, 2, …
    Here
    a2 – a1 = -6 – (-10)
    = 4
    a3 – a2 = -2 – (-6)
    = 4
    a4 – a3 = 2 – (-2)
    = 4
    As ak+1 – ak is the same throughout it is an AP with 1st term as -10 and common difference as 4.
    Thus the next three terms are
    a5 = a4 + d = 2 + 4 = 6
    a6 = a5 + d = 6 + 4 = 10
    a7 = a6 + d = 10 + 4 = 14
    (v) 3, 3 +, 3 + 2, 3 + 3, …
    Here a2 – a1 = 3 + -3 = 
    a3 – a2 = 3 + 2-(3+) = 
    a4 – a3 = 3 + 3-(3+2) = 
    In this problem, ak+1 –ak is the same throughout thus the given list is an AP with 1st term as 3 and common difference (d) as .
    Thus the next three terms are
    a5 = a4 + d = (3 + 3) +  = 3 + 4
    a6 = a5 + d = (3 + 4) +  = 3 +5
    a7 = a6 + d = (3 +5) +  = 3 + 6
 5.  Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) 0.2, 0.22, 0.222, 0.222, … (ii) 0, -4, -8, -12, … (iii), … (iv) 1, 3, 9, 27, … (v) a, 2a, 3a, 4a, …
  • Solution:(i) 0.2, 0.22, 0.222, 0.2222, …
    Here
    a2 – a1 = 0.22 – 0.2 = 0.02
    a3 – a2 = 0.222 – 0.22 = 0.002
    a4 – a3 = 0.2222 – 0.222 = 0.0002
    Hence ak+1 – ak is not the same throughout, hence it is not an AP.
    (ii) 0, -4, -8, -12
    Here
    a2 – a1 = -4 – 0 = -4
    a3 – a2 = -8 – (-4) = -4
    a4 – a3 = -12 – (-8) = -4
    Here ak+1 – ak is the same throughout hence it is an AP with 1st term = 0
    and common difference = -4
    Thus the next three terms are
    a5 = a4 + d = -12 + (-4) = -16
    a6 = a5 + d = -16 + (-4) = -20
    a7 = a6 + d = -20 + (-4) = -24
    (iii) , …
    Here
    a2 – a1 = - = 0
    a3 – a2 = –  = 0
    a4 – a3 = –  = 0
    Here ak+1 – ak = 0 is the same throughout hence the list is an AP with 1st term =  and common difference d = 0.
    Thus the next three terms are
    a5 = a4 + d =  + 0 =
    a6 = a5 + d =  + 0 = 
    a7 = a6 + d =  + 0 = .
    (iv) 1, 3, 9, 27, …
    Here the 1st term (a) = 1
    a2 – a1 = 3 – 1 = 2
    a3 – a2 = 9 - 3 = 6
    a4 – a3 = 27 – 9 = 18
    Since ak+1 – ak is not the same throughout this is not an AP.
    (v) a, 2a, 3a, 4a
    Here
    a2 – a1 = 2a - a = a
    a3 – a1 = 3a – 2a = a
    a4 – a3 = 4a – 3a = a
    Since ak+1 – a1 is the same throughout this is a AP with first term = a and common difference = a
    Thus the next three terms are
    a5 = a4 + d = 4a + a = 5a
    a6 = a5 + d = 5a + a = 6a
    a7 = a6 + d = 6a + a = 7a
 6.  Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) a, a2, a3, a4, … (ii) ,,, , …
  • Solution:(i) a, a2, a3, a4, …
    Here
    a2 – a1 = a2 - a = a(a – 1)
    a3 – a2 = a3 – a2 = a2(a – 1)
    a4 – a3 = a4 – a3 = a3(a – 1)
    Since ak+1 – ak is not the same throughout this is not an AP.
    (ii) ,,,, …
    Here
    a2 – a, =  -  = 2 -  =
    a3 – a2 =  -  = 3 - 2 = 
    a4 – a3 =  - = 4 - 3 = 
    Since ak+1 – ak is the same throughout it is an AP with the first term as  and common difference d = .
    Thus the next three terms are
    a5 = a4 + d = +  = 4 +  = 5
    a6 = a5 + d = 5 +  = 6
    a7 = a6 + d = 6 +  = 7.
    (iii)
    Here
    a2 – a1 =  = ( - 1)
    a3 – a2 =  = 3 -= (- )
    a4 – a3 = = 2 - 3 = (2 - ).
    Here ak+1 – ak is not the same throughout hence the list is not an AP.
    (iv) 12, 32, 52, 72
    Here
    a2 – a1 = 32 – 12 = 9 – 1 = 8
    a3 – a2 = 52 – 32 = 25 – 9 = 16
    a4 – a3 = 72 – 52 = 49 – 25 = 24
    Since ak+1 – ak is not the same throughout hence the list not an AP.
    (v) 12, 52, 72, 73, …
    Here
    a2 – a1 = 52 – 12 = 25 – 1 = 24
    a3 – a2 = 72 – 52 = 49 – 25 = 24
    a4 – a3 = 73 – 72 = 73 – 49 = 24
    Since ak+1 – ak = 24 is the same throughout this list is an AP with first term = 12 and common difference = 24.
    Thus the next three terms are
    a5 = a4 + d = 73 + 24 = 97
    a6 = a5 + d = 97 + 24 = 121
    a7 = a6 + d = 121 + 24 = 145.
 7.  Fill in the blanks in the following table, given that a is the first term, d the common difference and an the nth term of the AP:
 
a
d
n
an
(i)
(ii)
(iii)
(iv)
(v)
7
-18
….
-18.9
3.5
3
….
-3
2.5
0
8
10
18
…
105
...
0
-5
3.6
...

  • Solution:(i) a = 7, d = 3 , n = 8, an = ?
    an = a + (n – 1) ´ d
    = 7 + (8 - 1) ´ 3
    = 7 + 7 ´ 3 = 7 + 21 = 28.
    (ii) a = -18
    n = 10
    an = 0
    d = ?
    an = a + (n – 1) ´ d
    0 = -18 + (10 – 1) ´ d
    0 = -18 + 9 ´ d
    0 = -18 + 9d
    9d = 18
    d = 2.
    (iii) a = ?, d = -3, n = 18, an = -5
    an = a + (n – 1) ´ d
    -5 = a + (18 – 1) ´ (-3)
    -5 = a + 17 ´ (-3)
    -5 = a - 51
    a = 51 – 5 = 46.
    (iv) a = -18.9
    d = 2.5
    an = 3.6
    n = ?
    an = a + (n – 1) ´ d
    3.6 = -18.9 +(n – 1) ´ 2.5
    (n – 1) ´ 2.5 = 3.6 + 18.9
    (n – 1) =  =  ( To remove the decimal multiply and divide by 10)
    => n –1 = 9
    \ n = 10.
    (v) a = 3.5
    d = 0
    n = 105
    an = ?
    an = a + (n – 1) ´ d
    = 3.5 +(105 – 1) ´ 0
    an = 3.5.
 8.  Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10, 7, 4, …, is
(A) 97 (B) 77 (C) –77 (D) -87
(ii) 11th term of the AP: -3, , 2, …, is
  • Solution:(i) 30th term of the AP: 10, 7, 4, …, is
    nth term = a + (n – 1) ´ d, where n = 30
    30th term = 10 + (30 – 1) ´ (-3)
                  = 10 + (29 ´ - 3)
                  = 10 – 87 = - 77
    (C) is the correct answer.
    (ii) 11th term of the AP: -3,  , 2, …, is
    Here n = 11
    a = -3
    d =  – (-3)
      = = = 2
    an = a + (n – 1) ´ d
    an = -3+ (n – 1) ´ (2)
        = -3 +(11 – 1) ´  = -3 + 10 ´ 
        =– 3 + 25 = 22
    a13 = 22.
    (B) is the correct answer.
 9.  In the following APs, find the missing terms in the boxes:
(i) 2,  ,26 (ii)  , 13,  , 3 (iii) 5, , , 9
(iv) -4,, , , , 6    (v) , 38, , , , -22
  • Solution:(i) Let x be the missing term
    x - 2 = 26 - x
    2x = 28
    x = 14
    \ Missing term = 14
    (ii)  , 13,  , 3
    Let missing term in 1st box be x , missing term in the 2nd box be y,
    13 - x = y - 13 = 3 - y
    Now considering the last two ,
    y - 13 = 3 - y
    2y = 16
    y = 8.
    The value of y = 8.
    Here 13 - x = y - 13
    => 13 - x = 8 - 13
    x = 26 - 8
    = 18.
    (iii) 5, , , 9
    Here a = 5
    a4 = 9
    n = 4
    d = ?

    an = a + (n- 1) ´ d
    9 = 5 + (4 - 1) ´ d
     = 5 + 3d
    3d =  - 5 =  = 
    d =  = 
    \ If a1 = 5
    a2 = a1 + d = 5 + == = 6
    a3 = 6  +  = 8
    5 ,
    (iv) -4,  ,  ,  ,  , 6
    an = 6
    a = -4
    d = ?

    an = a + (n -1) ´ d
    6 = -4 + (6 -1) ´ d
    6 = -4 + 5d
    5d = 6 + 4 = 10
    d =  = 2
    \ -4,  ,  ,  ,  , 6 is the AP series.
    (v)  , 38,  ,  ,  , -22
    a2 = 38
    n = 6
    a6 = -22
    an = a + (n - 1) ´ d if n = 6,
    -22 = a + 5d ……. (1)
    if n = 2,
    38 = a + d ………. (2)

    - (2) => 4d = -60
    \ d = = -15.
    Thus a2 = a + d = a + (-15)
    38 = a - 15
    a = 38 + 15 = 53.
    a3 = 38 + (-15) = 23
    a4 = 23 + (-15) = 8
    a5 = 8 + (-15) = -7
    \ 53 , 38 , 23 , , , -22.
 10.  Which term of the AP : 3,8, 13, 18, …,is 78

Popular posts from this blog

worksheet 1 Triangles

1.    Give two different examples of pair of (i) similar figures. (ii) non-similar figures. Solution: Two square of sides 4 cm and 8 cm each. A rhombus and a trapezium .   2.    State whether the following quadrilaterals are similar or not      Solution: Similar   3.    In the figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). Solution: (i)   In  D ABC  DE is parallel to BC By Basic Proportionality Theorem  ------------------- (1) Given:  AD = 1.5 cm, DB = 3 cm, AE = 1 cm Let EC = ‘x’ cm Applying in (1) 1.5x = 3 x =  x = 2 cm EC = 2cm (ii) Since DE || BC, using BPT  …………………………. (1) Given: DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm Let AD be = x sub. in (1) x =  =  \  AD = 2.4 cm   4.    E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm,
Read more

plant cell

Read more

Types and Distribution of Forest and Wildlife Resources

Types and Distribution of Forest and Wildlife Resources In India, forest and wildlife resources are owned and looked after by the government of India through its Forest Department. Forests are classified under three categories by the Forest Department Reserved forests The most valuable forests are classified as Reserved forests. These forests are protected under the conservation acts. More than half of the total forest land in India comes under Reserved Forests. Protected Forests Forests that are protected from further destruction are called Protected Forests. One-third of the total forest areas in India are classified as Protected forests. Unclassed forests Forests and wastelands belonging to both government and private individuals and communities are classified as Unclassed Forests. Forest density map Madhya Pradesh has the largest area under  permanent forests . Andhra Pradesh, Uttarakhand, Kerala, Tamil Nadu, West Bengal, and M
Read more