TOPICS
Exercise - 5.3

Question-1 :-  Find the sum of the following APs:
(i) 2, 7, 12, ...., to 10 terms.
(ii) –37, –33, –29, ...., to 12 terms.
(iii) 0.6, 1.7, 2.8, ...., to 100 terms.
(iv) 1/15, 1/12, 1/10, ...., to 11 terms.

Solution :-
(i) Here, a = 2, d = 7 - 2 = 5, n = 10
    Sn = n/2 [2a + (n-1)d]
    S10 = 10/2 [2 x 2 + (10-1) x 5]
    S10 = 5[4 + 9 x 5]
    S10 = 5[4 + 45]
    S10 = 5 x 50
    S10 = 250
    
(ii) Here, a = -37, d =-33 + 37 = 4, n = 12
    Sn = n/2 [2a + (n-1)d]
    S12 = 12/2 [2 x (-37) + (12-1) x 4]
    S12 = 6[-74 + 11 x 4]
    S12 = 6[-74 + 44]
    S12 = 6 x (-30)
    S12 = -180
    
(iii) Here, a = 0.6, d = 1.7 - 0.6 = 1.1, n = 100
    Sn = n/2 [2a + (n-1)d]
    S100 = 100/2 [2 x 0.6 + (100-1) x 1.1]
    S100 = 50[1.2 + 99 x 1.1]
    S100 = 50[1.2 + 108.9]
    S100 = 50 x 110.1
    S100 = 5505.0
    
(iv) Here, a = 1/15, d =  1/12 - 1/15 = 1/60, n = 11
    Sn = n/2 [2a + (n-1)d]
    S11 = 11/2 [2 x 1/15 + (11-1) x 1/60]
    S11 = 11/2 [2/15 + 10 x 1/60]
    S11 = 11/2 [2/15 + 10/60]
    S11 = 11/2 x 18/60
    S11 = 99/60 = 33/20
    

Question-2 :-  Find the sums given below :
(i) 7 + 10 by 1/2 + 14 + .... + 84
(ii) 34 + 32 + 30 + . . . + 10
(iii) –5 + (–8) + (–11) + . . . + (–230)

Solution :-
(i) Here, a = 7, d = (10 by 1/2) - 7 = 21/2 - 7 = 7/2, l = 84
    For find n, By using formula
    l = a + (n-1)d
    84 = 7 + (n-1)(7/2)
    84 - 7 = 7n/2 - 7/2
    77 + 7/2 = 7n/2
    161/2 = 7n/2
    7n = 161
    n = 161/7 
    n = 23
    Now, using formula of sums
    Sn = n/2 [a + l]
    S23 = 23/2 [7 + 84]
    S23 = 23/2 x 91
    S23 = 2093/2
    
(ii) Here, a = 34, d = 32 - 34 = -2, l = 10
    For find n, By using formula
    l = a + (n-1)d
    10 = 34 + (n-1)(-2)
    10 - 34 = -2n + 2
    -24 - 2 = -2n
    -26 = -2n
    2n = 26
    n = 26/2 
    n = 13
    Now, using formula of sums
    Sn = n/2 [a + l]
    S13 = 13/2 [34 + 10]
    S13 = 13/2 x 44
    S13 = 13 x 22 
    S13= 286
    
(iii) Here, a = -5, d = -8 - (-5) = -8 + 5 = -3, l = -230
    For find n, By using formula
    l = a + (n-1)d
    -230 = -5 + (n-1)(-3)
    -230 + 5 = -3n + 3
    -225 - 3 = -3n
    -228 = -3n
    3n = 228
    n = 228/3 
    n = 76
    Now, using formula of sums
    Sn = n/2 [a + l]
    S76 = 76/2 [-5 + (-230)]
    S76 = 76/2 x (-235)
    S76 = 38 x (-235) 
    S76= -8930
    

Question-3 :-  In an AP:
(i) given a = 5, d = 3, an = 50, find n and Sn.
(ii) given a = 7, a13 = 35, find d and S13.
(iii) given a12 = 37, d = 3, find a and S12.
(iv) given a3 = 15, S10 = 125, find d and a10.
(v) given d = 5, S9 = 75, find a and a9.
(vi) given a = 2, d = 8, Sn = 90, find n and an.
(vii) given a = 8, an = 62, Sn = 210, find n and d.
(viii) given an = 4, d = 2, Sn = –14, find n and a.
(ix) given a = 3, n = 8, S = 192, find d.
(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Solution :-
(i) Given : a = 5, d = 3, an = 50, n = ?, Sn = ?
    For find n, By using forula
    an = l = a + (n-1)d
    50 = 5 + (n-1)3
    50 - 5 = 3n - 3
    45 + 3 = 3n
    48 = 3n
    n = 48/3
    n = 16
    Now, For find Sn By using formula
    Sn = n/2 [a + l]
    S16 = 16/2 [5 + 50]
    S16 = 8 x 55
    S16 = 440 
    
(ii) Given : a = 7, a13 = 35, d = ?,  S13 = ?
    For find d, By using forula
    an = l = a + (n-1)d
    a13 = 7 + (13-1)d
    35 = 7 + 12d
    12d = 35 - 7
    12d = 28
    d = 28/12
    d = 7/3
    Now, For find Sn By using formula
    Sn = n/2 [a + l]
    S13 = 13/2 [7 + 35]
    S13 = 13/2 x 42
    S13 = 13 x 21
    S13 = 273
    
(iii) Given : a12 = 37, d = 3, a = ?, S12 = ?
    For find d, By using forula
    an = l = a + (n-1)d
    a12 = a + (12-1)3
    37 = a + 11 x 3
    37 = a + 33
    a = 37 - 33
    a = 4
    Now, For find Sn By using formula
    Sn = n/2 [a + l]
    S12 = 12/2 [4 + 37]
    S12 = 12/2 x 41
    S12 = 6 x 41
    S12 = 246   
    
(iv) Given : a3 = 15, S10 = 125, d = ?, a10 = ?
    a3 = a + (3-1)d
    15 = a + 2d
    a + 2d = 15 ....(i)
    S10 = n/2 [2a + (n-1)d]
    125 = 10/2 [2a + (10-1)d]
    125 = 5 [2a + 9d]
    125 = 10a + 45d
    By taking out 5 from both sides, we get
    25 = 2a + 9d
    2a + 9d = 25 ....(ii)

    By substracting (ii) to (i) equation and also multiply by 2 (i) equation 
    2a + 9d - 2a - 4d = 25 - 30
    5d = -5
    d = -5/5
    d = -1
    Put the value of d = -1 in (i) equation
    a + 2 x (-1) = 15
    a - 2 = 15
    a = 15 + 2
    a = 17 

    a10 = 17 + (10-1)(-1)
    a10 = 17 + 9 x (-1)
    a10 = 17 - 9
    a10 = 8
    
(v) Given : d = 5, S9 = 75, a = ?, a9 = ?
    S9 = 9/2 [2a + (9-1)5]
    75 = 9/2 [2a + 8 x 5]
    150 = 9[2a + 40]
    150 = 18a + 360
    18a = 150 - 360
    18a = -210
    a = -210/18
    a = -35/3
    Now, for a9
    a9 = -35/3 + (9-1)5
    a9 = -35/3 + 40
    a9 = 85/3
    
(vi) Given : a = 2, d = 8, Sn = 90, n = ?,  an = ?
    Sn = n/2 [2a + (n-1)d]
    90 = n/2 [2 x 2 + (n-1)8]
    180 = n[4 + 8n - 8]
    180 = n[8n - 4]
    180 = 8n2 - 4n
    Taking out 4 from both sides, we get
    45 = 2n2 - n
    2n2 - n - 45 = 0
    2n2 - 10n + 9n - 45 = 0
    2n(n - 5) + 9(n - 5) = 0
    (2n + 9)(n - 5) = 0
    2n + 9 = 0, n - 5 = 0
    2n = -9, n = 5
    n = -9/2 (Rejected), n = 5
    n cannot be -9/2. As the number of terms can neither be negative nor fractional, therefore, n = 5.
    Now, for an
    an = a + (n-1)d
    a5 = 2 + (5-1)8
    a5 = 2 + 32
    a5 = 34
    
(vii) Given : a = 8, an = 62, Sn = 210, n = ?,  d = ?
    We know that an = l, By using formula
    Sn = n/2 [a + l]
    210 = n/2 [8 + 62]
    210 = n/2 x 70
    210 = 35n
    n = 210/35
    n = 6
    Now, using formula for find d
    an = a + (n-1)d
    62 = 8 + (6-1)d
    62 - 8 = 5d
    54 = 5d
    d = 54/5
    
(viii) Given : an = 4, d = 2, Sn = –14, n = ?, a = ?
    an = a + (n-1)d
    4 = a + (n-1)2
    4 = a + 2n - 2
    4 + 2 = a + 2n
    a + 2n = 6
    a = 6 - 2n ....(1)

    Sn = n/2 [a + l]
    -14 = n/2 [a + 4] 
    -28 = na + 4n
    -28 = n(6 - 2n) + 4n       [By Using (1)]
    -28 = 6n - 2n2 + 4n
    -28 = 10n - 2n2
    2n2 - 10n - 28 = 0
    By taking out 2
    2(n2 - 5n - 14) = 0
    n2 - 5n - 14 = 0
    n2 - 7n + 2n - 14 = 0
    n(n - 7) + 2(n - 7) = 0
    (n + 2)(n - 7) = 0
    n + 2 = 0, n - 7 = 0
    n = -2 (Rejected), n = 7
    n cannot be -2. As the number of terms can neither be negative nor fractional, therefore, n = 7.
    Put the value of n = 7 in (1) equation
    a = 6 - 2 x 7
    a = 6 - 14
    a = -8
    
(ix) Given : a = 3, n = 8, S = 192, d = ?
    Sn = n/2 [2a + (n-1)d]
    192 = 8/2 [2 x 3 + (8-1)d]
    192 = 4[6 + 7d]
    192/4 = 6 + 7d
    48 - 6 = 7d
    42 = 7d
    d = 42/7
    d = 6
    
(x) Given : l = 28, S = 144, n = 9, a = ?
    Sn = n/2 [a + l]
    144 = 9/2 [a + 28]
    288/9 = a + 28
    32 = a + 28
    a = 32 - 28
    a = 4
    

Question-4 :-  How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?

Solution :-
    Here, a = 9, d = 17 - 9 = 8, Sn = 636
    Sn = n/2 [2a + (n-1)d]
    636 = n/2 [2 x 9 + (n-1)8]
    1272 = n[18 + 8n - 8]
    1272 = n[10 + 8n]
    1272 = 10n + 8n2
    Taking out 2 from both sides, we get
    636 = 5n + 4n2
    4n2 + 5n - 636 = 0
    4n2 + 53n − 48n − 636 = 0
    n(4n + 53) − 12(4n + 53) = 0 
    (4n + 53)(n − 12) = 0
    4n + 53 = 0 or n − 12 = 0
    n = -53/4(Rejected), n = 12
    n cannot be -53/4. As the number of terms can neither be negative nor fractional, therefore, n = 12.
    

Question-5 :-  The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution :-
    Here, a = 5, l = 45, Sn = 400, n = ?, d = ?
    Sn = n/2 [a + l]
    400 = n/2 [5 + 45]
    800 = n x 50
    n = 800/50
    n = 16
    Now, l = a + (n-1)d
    45 = 5 + (16-1)d
    45 - 5 = 15d
    40 = 15d
    d = 40/15
    d = 8/3
    

Question-6 :-  The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution :-
    Here, a = 17, l = 350, d = 9, n = ?, Sn = ?
    l = a + (n-1)d
    350 = 17 + (n-1)9
    350 - 17 = 9n - 9
    333 + 9 = 9n
    342 = 9n
    n = 342/9
    n = 38
    Now, Sn = n/2 [a + l]
    S38 = 38/2 [17 + 350]
    S38 = 19 x 367
    S38 = 6973
    

Question-7 :-  Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution :-
    Here, n = 22, d = 7, a22 = 149, a = ?, S22 = ?
    a22 = a + 21d = 149
    a + 21 x 7 = 149
    a + 147 = 149
    a = 149 - 147
    a = 2
    Now, Sn = n/2 [a + l]
    S22 = 22/2 [2 + 149]
    S22 = 11 x 151
    S22 = 1661
    

Question-8 :-  Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution :-
    Here, n = 51, S51 = ?
    a2 = a + d = 14 ....(1)
    a3 = a + 2d = 18 ....(2)
    Substracting (2) to (1) equation, we get
    a + 2d - a - d = 18 - 14
    d = 4
    Put the value of d = 4 in (1) equation
    a + 4 = 14
    a = 14 - 4 
    a = 10
    Now, Sn = n/2 [2a + (n-1)d]
    S51 = 51/2 [2 x 10 + (51-1)4]
    S51 = 51/2 [20 + 200]
    S51 = 51/2 x 220
    S51 = 51 x 110
    S51 = 5610
    

Question-9 :-  If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution :-
    Here, S7 = 49, S17 = 289, Sn = ?
    S7 = 49
    7/2 [2a + (7-1)d] = 49,
    2a + 6d = (49 x 2)/7
    2a + 6d = 14
    a + 3d = 7 ....(1)
    S17 = 289
    17/2 [2a + (17-1)d] = 289
    2a + 16d = (289 x 2)/17
    2a + 16d = 34
    a + 8d = 17 ....(2)
    Substracting (2) to (1) equation, we get
    a + 8d - a - 3d = 17 - 7
    5d = 10
    d = 10/5
    d = 2
    Put the value of d = 2 in (1) equation
    a + 3 x 2 = 7
    a + 6 = 7
    a = 7 - 6
    a = 1
    Now, Sn = n/2 [2a + (n-1)d]
    Sn = n/2 [2 x 1 + (n-1)2]
    Sn = n/2 [2 + 2n - 2]
    Sn = n/2 x 2n
    Sn = n2
    

Question-10 :-  Show that a1, a2, . . ., an, . . . form an AP where an is defined as below :
(i) an = 3 + 4n
(ii) an = 9 – 5n
Also find the sum of the first 15 terms in each case.

Solution :-
(i) an = 3 + 4n
    a1 = 3 + 4 x 1 = 3 + 4 = 7
    a2 = 3 + 4 x 2 = 3 + 8 = 11
    a3 = 3 + 4 x 3 = 3 + 12 = 15
    Therefore, A.P. is 7, 11, 15 .....
    Now, n = 15, a = 7, d = 11 - 7 = 4
    Sn = n/2 [2a + (n-1)d]
    S15 = 15/2 [2 x 7 + (15-1)4]
    S15 = 15/2 [14 + 14 x 4]
    S15 = 15/2 [14 + 56]
    S15 = 15/2 x 70
    S15 = 15 x 35
    S15 = 525
    
(ii) an = 9 - 5n
    a1 = 9 - 5 x 1 = 9 - 5 = 4
    a2 = 9 - 5 x 2 = 9 - 10 = -1
    a3 = 9 - 5 x 3 = 9 - 15 = -6
    Therefore, A.P. is 4, -1, -6 .....
    Now, n = 15, a = 4, d = -1 - 4 = -5
    Sn = n/2 [2a + (n-1)d]
    S15 = 15/2 [2 x 4 + (15-1)(-5)]
    S15 = 15/2 [8 + 14 x (-5)]
    S15 = 15/2 [8 - 70]
    S15 = 15/2 x (-62)
    S15 = 15 x (-31)
    S15 = -465
    

Question-11 :-  If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution :-
    Here, Sn = 4n − n2
    First term, a = S1 = 4(1) − (1)2 = 4 − 1 = 3 
    Sum of first two terms = S2 = 4(2) − (2)2 = 8 − 4 = 4
    Second term, a2 = S2 − S1 = 4 − 3 = 1 
    d = a2 − a = 1 − 3 = −2
    an = a + (n−1)d = 3 + (n−1)(−2) = 3 − 2n + 2 = 5 − 2n
    Therefore, a3 = 5 − 2(3) = 5 − 6 = −1 
    a10 = 5 − 2(10) = 5 − 20 = −15
    Hence, the sum of first two terms is 4. 
    The second term is 1. 
    3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively.
    

Question-12 :-  Find the sum of the first 40 positive integers divisible by 6.

Solution :-
    Here, n = 40
    Divisible by 6 values are 6, 12, 18, .....
    a = 6, d = 12 - 6 = 6
    Sn = n/2 [2a + (n-1)d]
    S40 = 40/2 [2 x 6 + (40-1)6]
    S40 = 20 [12 + 39 x 6]
    S40 = 20 [12 + 234]
    S40 = 20 x 246
    S40 = 4920
    

Question-13 :-  Find the sum of the first 15 multiples of 8.

Solution :-
    Here, n = 15
    Multiples of 8 values are 8, 16, 24, .....
    a = 8, d = 16 - 8 = 8
    Sn = n/2 [2a + (n-1)d]
    S40 = 15/2 [2 x 8 + (15-1)8]
    S40 = 15/2 [16 + 14 x 8]
    S40 = 15/2 [16 + 112]
    S40 = 15/2 x 128
    S40 = 15 x 64
    S40 = 960
    

Question-14 :-  Find the sum of the odd numbers between 0 and 50.

Solution :-
    Odd numbers between 0 and 50 are 1, 3, 5, ....49
    a = 1, d = 3 - 1 = 2, l = 49
    l = a + (n-1)d
    49 = 1 + (n-1)2
    49 - 1 = 2n - 2
    48 + 2 = 2n
    50 = 2n
    n = 50/2
    n = 25
    Now, Sn = n/2 [a + l]
    S25 = 25/2 [1 + 49]
    S25 = 25/2 x 50
    S25 = 25 x 25
    S25 = 625
    

Question-15 :-  A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days?

Solution :-
    It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
    a = 200, d = 50
    Penalty that has to be paid if he has delayed the work by 30 days
    S30 = 30/2 [2 x 200 + (30-1)50]
    S30 = 15[400 + 29 x 50]
    S30 = 15[400 + 1450]
    S30 = 15 x 1850
    S30 = 27750
    Therefore, the contractor has to pay ₹27750 as penalty. 
    

Question-16 :-  A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Solution :-
    Let the cost of 1st prize be P. 
    Cost of 2nd prize = P − 20 
    Cost of 3rd prize = P − 40
    It can be observed that the cost of these prizes are in an A.P. having common difference as −20 and first term as P.
    a = P, d = −20
    Given that, S7 = 700
    S7 = 7/2 [2 x P + (7-1)(-20)]
    700 = 7/2 [2P - 120]
    1400/7 = 2P - 120
    200 + 120 = 2P
    2P = 320
    P = 320/2
    P = 160
    a = 160
    Therefore, the value of each of the prizes was ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, and ₹40.  
    

Question-17 :-  In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution :-
    It can be observed that the number of trees planted by the students is in an AP.
    1, 2, 3, 4, 5,...... 12 
    a = 1, d = 2 − 1 = 1, n = 12
    Sn = n/2 [2a + (n-1)d]
    S12 = 12/2 [2 x 1 + (12-1)1]
    S12 = 6[2 + 11]
    S12 = 6 x 13
    S12 = 78

    Therefore, number of trees planted by 1 section of the classes = 78
    Number of trees planted by 3 sections of the classes = 3 × 78 = 234
    Therefore, 234 trees will be planted by the students. 
    

Question-18 :-  A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre atA, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)

Solution :-
    Semi-perimeter of circle = πr
    I1 = π(0.5) = π/2 cm
    I2 = π(1) = π cm
    I3 = π(1.5) = 3π/2 cm
    Therefore, I1, I2, I3 ,i.e. the lengths of the semi-circles are in an A.P.,
    π/2, π, 3π/2, .....
    a = π/2, d = π - π/2 = π/2, n = 13, S13 = ?
    S13 = 13/2 [2 x π/2 + (13-1)(π/2)]
    S13 = 13/2 [π + 12 x π/2]
    S13 = 13/2 [π + 6π]
    S13 = 13/2 x 7π
    S13 = 13/2 x 7 x 22/7
    S13 = 13 x 11
    S13 = 143
    Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm. 
    

Question-19 :-  200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see Figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Solution :-
    It can be observed that the numbers of logs in rows are in an A.P. 20, 19, 18, ....
    For this A.P., a = 20, d = a2 − a1 = 19 − 20 = −1
    Let a total of 200 logs be placed in n rows. 
    Sn = 200
    n/2 [2 x 20 + (n-1)(-1)] = 200
    n[40 - n + 1] = 400
    n[-n + 41] = 400
    -n2 + 41n - 400 = 0
    Taking out (-1), we get
    n2 − 41n + 400 = 0
    n2 − 16n − 25n + 400 = 0
    n(n − 16) − 25(n − 16) = 0 
    (n − 16)(n − 25) = 0
    Either (n − 16) = 0 or n − 25 = 0 
    n = 16 or n = 25
    an = a + (n−1)d
    a16 = 20 + (16 − 1)(−1) 
    a16 = 20 − 15
    a16 = 5 
    Similarly,
    a25 = 20 + (25−1)(−1) 
    a25 = 20 − 24
    a25 = −4

    Clearly, the number of logs in 16th row is 5. 
    However, the number of logs in 25th row is negative, which is not possible.
    Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.
    

Question-20 :-  In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see Figure). A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution :-
    The distances of potatoes are as follows.
    5, 8, 11, 14, ....
    It can be observed that these distances are in A.P.
    a = 5, d = 8 − 5 = 3, n = 10
    Sn = n/2 [2a + (n-1)d]
    S10 = 10/2 [2 x 5 + (10-1)3]
    S10 = 5[10 + 27]
    S10 = 5 x 37
    S10 = 185
    As every time she has to run back to the bucket, therefore, 
    the total distance that the competitor has to run will be two times of it.
    Therefore, total distance that the competitor will run = 2 × 185 = 370 m
    
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