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

Acids, Bases and salts extra qus

1.    What is an acid? Solution: An acid is a hydrogen-containing chemical compound which, when dissolved in water, gives hydrogen ion (H + ) or hydrated hydrogen ion (H 2 O. H + ) or hydronium ion (H 3 O + ).   2.    What are bases and alkalies? Solution: Oxides and hydroxides of metals and metal like radicals (e.g., NH4 +  ions) are called bases. Bases ionise to give OH -  ions in aqueous solution. Bases may be soluble or insoluble in water. The soluble bases are called alkalies. Thus all alkalies are bases but all bases are not alkalies. Examples NaOH and Cu (OH) 2  both are bases, but, since NaOH is soluble in water, it is an alkali. On the other hand, since Cu (OH) 2  is insoluble in water, it is not an alkali. Other examples of alkalies are KOH, Ca (OH) 2  and NH 4 OH.   3.    Define pH. Solution: pH of a given solution is the negative logarithm to the base 10 of the hydrogen ion concentration, [H +...
Read more

Project Tiger

Project Tiger  is a wildlife conservation movement initiated in  India  in 1973 to protect  tigers . The project aims at tiger conservation in specially constituted  tiger reserves  representative of various regions throughout India and strives to maintain viable populations of Bengal tigers in their natural environment. In 2008 there were more than 40 Project Tiger reserves covering an area over 37,761 km 2  (14,580 sq mi). Project Tiger helped to increase the population of these tigers from 1,200 in the 1970s to 3,500 in 1990s. However, a 2008 census held by the Government of India revealed that the tiger population had dropped to 1,411. Since then the government has pledged US$153 million to further fund the project, set-up a Tiger Protection Force to combat  poachers , and fund the relocation of up to 200,000 villagers to minimize human-tiger conflicts. The number of tigers in India's wild has gone up by 20%, according to...
Read more

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 ...
Read more