TOPICS
Exercise - 5.1

Question-1 :-  In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8 % per annum.

Solution :-
(i) Here, According to Question
  Taxi fare for 1st km = ₹15
  Taxi fare for 1st 2 km = 15 + 8 = ₹23 
  Taxi fare for 1st 3 km = 23 + 8 = ₹31 
  Taxi fare for 1st 4 km = 31 + 8 = ₹39
  Therefore, 15, 23, 31, 39 .... forms an A.P. because every term is 8 more than the preceding term.
    
(ii) Let the initial volume of air present in a cylinder be V litre. 
  In each stroke, the vacuum pump removes 1/4 of air remaining in the cylinder at a time.
  In other words, after every stroke, only 1 - 1/4 = 3/4 th  part of air will remain.
  Therefore, volumes will be V, (3V/4), (3V/4)2, (3V/4)3 ....
  So, it can be observed that the adjacent terms of this series do not have the same difference between them. 
  Therefore, this is not an A.P.
    
(iii) Here, According to Question
  Cost of digging for 1st metre = ₹150
  Cost of digging for 1st 2 metres = 150 + 50 = ₹200
  Cost of digging for 1st 3 metres = 200 + 50 = ₹250
  Cost of digging for 1st 4 metres = 250 + 50 = ₹300
  Therefore, 150, 200, 250, 300 .... forms an A.P. because every term is 50 more than the preceding term.
    
(iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be P(1+r/100)n after n years.
  Therefore, after every year, our money will be 1000(1+0.08), 1000(1+0.08)2, 1000(1+0.08)3, 1000(1+0.08)4 ....
  So, adjacent terms of this series do not have the same difference between them. 
  Therefore, this is not an A.P. 
    

Question-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 = 1/2
(v) a = –1.25, d = –0.25

Solution :-
(i) Given : a = 10, d = 10
  an = a + (n-1)d
  n = 1, 2, 3, 4
  a1 = 10 + (1-1)10 = 10 + 0 = 10
  a2 = 10 + (2-1)10 = 10 + 10 = 20
  a3 = 10 + (3-1)10 = 10 + 20 = 30
  a4 = 10 + (4-1)10 = 10 + 30 = 40
  So, Four terms of A.P. is 10, 20, 30 and 40
    
(ii) Given : a = -2, d = 0
  an = a + (n-1)d
  n = 1, 2, 3, 4
  a1 = -2 + (1-1)0 = -2 + 0 = -2
  a2 = -2 + (2-1)0 = -2 + 0 = -2
  a3 = -2 + (3-1)0 = -2 + 0 = -2
  a4 = -2 + (4-1)0 = -2 + 0 = -2
  So, Four terms of A.P. is -2, -2, -2 and -2
    
(iii) Given : a = 4, d = -3
  an = a + (n-1)d
  n = 1, 2, 3, 4
  a1 = 4 + (1-1)(-3) = 4 + 0 = 4
  a2 = 4 + (2-1)(-3) = 4 - 3 = 1
  a3 = 4 + (3-1)(-3) = 4 - 6 = -2
  a4 = 4 + (4-1)(-3) = 4 - 9 = -5
  So, Four terms of A.P. is 4, 1, -2 and -5
    
(iv) Given : a = -1, d = 1/2
  an = a + (n-1)d
  n = 1, 2, 3, 4
  a1 = -1 + (1-1)1/2 = -1 + 0 = -1
  a2 = -1 + (2-1)1/2 = -1 + 10 = -1/2
  a3 = -1 + (3-1)1/2 = -1 + 20 = 0
  a4 = -1 + (4-1)1/2 = -1 + 30 = 1/2
  So, Four terms of A.P. is -1, -1/2, 0 and 1/2
    
(v) Given : a = -1.25, d = -0.25
  an = a + (n-1)d
  n = 1, 2, 3, 4
  a1 = -1.25 + (1-1)(-0.25) = -1.25 + 0 = -1.25
  a2 = -1.25 + (2-1)(-0.25) = -1.25 - 0.25 = -1.50
  a3 = -1.25 + (3-1)(-0.25) = -1.25 - 0.50 = -1.75
  a4 = -1.25 + (4-1)(-0.25) = -1.25 - 0.75 = -2.00
  So, Four terms of A.P. is -1.25, -1.50, -1.75 and -2.00 
    

Question-3 :-  For the following APs, write the first term and the common difference:
(i) 3, 1, –1, –3, ....
(ii) –5, –1, 3, 7, ....
(iii) 1/3, 5/3, 9/3, 13/3 ....
(iv) 0.6, 1.7, 2.8, 3.9, ....

Solution :-
(i) 3, 1, –1, –3, ....
  Here, First Term (a) = 3
  and Common difference (d) = a2 - a1 = 1 - 3 = -2
    
(ii) –5, –1, 3, 7, ....
  Here, First Term (a) = -5
  and Common difference (d) = a2 - a1 = -1 - (-5) = -1 + 5 = 4
    
(iii) 1/3, 5/3, 9/3, 13/3 ....
  Here, First Term (a) = 1/3
  and Common difference (d) = a2 - a1 = 5/3 - 1/3 = 4/3
    
(iv) 0.6, 1.7, 2.8, 3.9, ....
  Here, First Term (a) = 0.6
  and Common difference (d) = a2 - a1 = 1.7 - 0.6 = 1.1
    

Question-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, 5/2, 3, 7/2, ....
(iii) -1.2, -3.2, -5.2, -7.2, ....
(iv) -10, -6, -2, 2, ....
(v) 3, 3+√2, 3+2√2, 3+3√2, ....
(vi) 0.2, 0.22, 0.222, 0.2222, ....
(vii) 0, -4, -8, -12, ....
(viii) -1/2, -1/2, -1/2, -1/2, ....
(ix) 1, 3, 9, 27, ....
(x) a, 2a, 3a, 4a, ....
(xi) a, a2, a3, a4, ....
(xii) √2, √8, √18, √32, ....
(xiii) √3, √6, √9, √12, ....
(xiv) 12, 32, 52, 72, ....
(xv) 12, 52, 72, 73, ....

Solution :-
(i) 2, 4, 8, 16, ....
  d1 = a2 - a1 = 4 - 2 = 2,
  d2 = a3 - a2 = 8 - 4 = 4
  Here, d1 ≠ d2 
  So, This series is not form in A.P.
    
(ii) 2, 5/2, 3, 7/2, ....
  d1 = a2 - a1 = 5/2 - 2 = 1/2,
  d2 = a3 - a2 = 3 - 5/2 = 1/2
  Here, d1 = d2 
  So, This series is form in A.P.
  Now, a = 2, d = 1/2
  According to question, 3 more terms are a5, a6 and a7
  a5 = 2 + (5-1)1/2 = 2 + 2 = 4
  a6 = 2 + (6-1)1/2 = 2 + 5/2 = 9/2 
  a7 = 2 + (7-1)1/2 = 2 + 3 = 5
    
(iii) -1.2, -3.2, -5.2, -7.2, ....
  d1 = a2 - a1 = -3.2 + 1.2 = -2.0,
  d2 = a3 - a2 = -5.2 + 3.2 = -2.0
  Here, d1 = d2 
  So, This series is form in A.P.
  Now, a = -1.2, d = -2.0
  According to question, 3 more terms are a5, a6 and a7
  a5 = -1.2 + (5-1)(-2.0) = -1.2 - 8 = 9.2
  a6 = -1.2 + (6-1)(-2.0) = -1.2 - 10 = 11.2
  a7 = -1.2 + (7-1)(-2.0) = -1.2 - 12 = 13.2
    
(iv) -10, -6, -2, 2, .... 
  d1 = a2 - a1 = -6 + 10 = 4,
  d2 = a3 - a2 = -2 + 6 = 4
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = -10, d = 4
  According to question, 3 more terms are a5, a6 and a7
  a5 = -10 + (5-1)(4) = -10 + 16 = 6
  a6 = -10 + (6-1)(4) = -10 + 20 = 10
  a7 = -10 + (7-1)(4) = -10 + 24 = 14 
    
(v) 3, 3+√2, 3+2√2, 3+3√2, ....
  d1 = a2 - a1 = 3+√2 - 3 = √2
  d2 = a3 - a2 = 3+2√2 - 3-√2 = √2
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = 3, d = √2
  According to question, 3 more terms are a5, a6 and a7
  a5 = 3 + (5-1)(√2) = 3+4√2
  a6 = 3 + (6-1)(√2) = 3+5√2
  a7 = 3 + (7-1)(√2) = 3+6√2 
    
(vi) 0.2, 0.22, 0.222, 0.2222, .... 
  d1 = a2 - a1 = 0.22 - 0.2 = 0.02
  d2 = a3 - a2 = 0.222 - 0.22 = 0.002
  Here, d1 ≠ d2 
  So, This series is not form in A.P.
    
(vii) 0, -4, -8, -12, ....
  d1 = a2 - a1 = -4 - 0 = -4
  d2 = a3 - a2 = -8 + 4 = -4
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = 0, d = -4
  According to question, 3 more terms are a5, a6 and a7
  a5 = 0 + (5-1)(-4) = 0 - 16 = -6
  a6 = 0 + (6-1)(-4) = 0 - 20 = -10
  a7 = 0 + (7-1)(-4) = 0 - 24 = -14 
    
(viii) -1/2, -1/2, -1/2, -1/2, ....
  d1 = a2 - a1 = -1/2 + 1/2 = 0
  d2 = a3 - a2 = -1/2 + 1/2 = 0
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = -1/2, d = 0
  According to question, 3 more terms are a5, a6 and a7
  a5 = -1/2 + (5-1)(0) = -1/2 - 0 = -1/2
  a6 = -1/2 + (6-1)(0) = -1/2 - 0 = -1/2
  a7 = -1/2 + (7-1)(0) = -1/2 - 0 = -1/2 
    
(xi) 1, 3, 9, 27, ....
  d1 = a2 - a1 = 3 - 1 = 2
  d2 = a3 - a2 = 9 - 3 = 6
  Here, d1 ≠ d2 
  So, This series is not form in A.P.
    
(x) a, 2a, 3a, 4a, ....
  d1 = a2 - a1 = 2a - a = a
  d2 = a3 - a2 = 3a - 2a = a
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = a, d = a
  According to question, 3 more terms are a5, a6 and a7
  a5 = a + (5-1)(a) = a + 4a = 5a
  a6 = a + (6-1)(a) = a + 5a = 6a
  a7 = a + (7-1)(a) = a + 6a = 7a 
    
(xi) a, a2, a3, a4, ....
  d1 = a2 - a1 = a2 - a = a(a-1)
  d2 = a3 - a2 = a3 - a2 = a2(a-1)
  Here, d1 ≠ d2 
  So, This series is not form in A.P.  
    
(xii) √2, √8, √18, √32, ....
  d1 = a2 - a1 = √8 - √2 = 2√2 - √2 = √2
  d2 = a3 - a2 = √18 - √8 = 3√2 - 2√2 = √2
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = √2, d = √2
  According to question, 3 more terms are a5, a6 and a7
  a5 = √2 + (5-1)(√2) = √2 + 4√2 = 5√2 = √50
  a6 = √2 + (6-1)(√2) = √2 + 5√2 = 6√2 = √72
  a7 = √2 + (7-1)(√2) = √2 + 6√2 = 7√2 = √98
    
(xiii) √3, √6, √9, √12, .... 
  d1 = a2 - a1 = √6 - √3
  d2 = a3 - a2 = √9 - √6
  Here, d1 ≠ d2 
  So, This series is not form in A.P.  
    
(xiv) 12, 32, 52, 72, .... 
  d1 = a2 - a1 = 32 - 12 = 9 - 1 = 8
  d2 = a3 - a2 = 52 - 32 = 25 - 9 = 16
  Here, d1 ≠ d2 
  So, This series is not form in A.P. 
    
(xv) 12, 52, 72, 73, ....
  d1 = a2 - a1 = 52 - 12 = 25 - 1 = 24
  d2 = a3 - a2 = 72 - 52 = 49 - 25 = 24
  Here, d1 = d2 
  So, This series is form in A.P. 
  Now, a = 1, d = 24
  According to question, 3 more terms are a5, a6 and a7
  a5 = 1 + (5-1)(24) = 1 + 96 = 97
  a6 = 1 + (6-1)(24) = 1 + 120 = 121
  a7 = 1 + (7-1)(24) = 1 + 144 = 145 
    
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