TOPICS
Unit-5(Examples)

Example-1 :-  For the AP : 3/2, 1/2, -1/2, -3/2 . . ., write the first term a and the common difference d.

Solution :-
  First Term (a) = 3/2
  Common Difference (d) = 1/2 - 3/2 = -1
    

Example-2 :-  Which of the following list of numbers form an AP? If they form an AP, write the next two terms :
(i) 4, 10, 16, 22, . . .
(ii) 1, – 1, – 3, – 5, . . .
(iii) – 2, 2, – 2, 2, – 2, . . .
(iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . .

Solution :-
(i) We have a2 – a1 = 10 – 4 = 6 
            a3 – a2 = 16 – 10 =  6 
            a4 – a3 = 22 – 16  = 6 
    i.e., an + 1 – an  is the same every time. 
    So, the given list of numbers forms an AP with the common difference d = 6. 
    The next two terms are: 22 + 6 = 28 and 28 + 6 = 34. 
    
(ii) We have a2 – a1 =  – 1 – 1 = – 2 
             a3 – a2 =  – 3 – ( –1 ) = – 3 + 1 = – 2 
             a4 – a3 = – 5 – ( –3 ) = – 5 + 3 = – 2 
    i.e., an + 1 – an  is the same every time. 
    So, the given list of numbers forms an AP with the common difference d = –2. 
    The next two terms are: – 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9  
    
(iii) We have a2 – a1 = 2 – (– 2) = 2 + 2 = 4 
              a3 – a2  = – 2 – 2 = – 4 
    As a2 – a1  ≠ a3 – a2. 
    So, the given list of numbers does not form an AP.
    
(iv) We have a2 – a1 = 1 – 1 = 0 
             a3 – a2 = 1 – 1 = 0 
             a4 – a3 = 2 – 1 = 1 
    Here, a2 – a1 = a3 – a2 ≠ a4 – a3. 
    So, the given list of numbers does not form an AP.
    

Example-3 :-  Find the 10th term of the AP : 2, 7, 12, . . .

Solution :-
  Here, a = 2, d = 7 - 2 = 5, n = 10
  Now, an = a + (n-1)d
  a10 = 2 + (10-1) x 5 = 2 + 45 = 47
  Therefore, the 10th term of the given AP is 47.
    

Example-4 :-  Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer.

Solution :-
  We have, l = -81, a = 21, d = 18 - 21 = -3
  Now, a + (n-1)d = an
  21 + (n-1) x (-3) = -81
  -3n + 3 = -81 - 21
  -3n = -81 - 21 - 3
  -3n = -105
  n = 105/3
  n = 35 
  Therefore, the 35th term of the given AP is – 81. 
  Next, we want to know if there is any n for which an = 0. If such an n is there, then 
  21 + (n-1) x (-3) = 0
  -3n + 3 = -21
  -3n = -21 - 3
  -3n = -24
  n = 24/3
  n = 8
  So, the eighth term is 0.
    

Example-5 :-  Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution :-
  Here, 3rd term = a3 = a + 2d = 5 .....(i)
        7th term = a7 = a + 6d = 9 .....(ii)
  Now, solving equation (i) & (ii)
  a + 2d - a - 6d = 5 - 9
  -4d = -4
  d = 1
  Putting value d = 1 in equation (i), 
  a + 2 x 1 = 5
  a = 5 - 2
  a = 3
  Therefore, General Form of an AP = a, a + d, a + 2d, ........
  AP = 3, 3 + 1, 3 + 2 x 1, .....
  AP = 3, 4, 5, .....
    

Example-6 :-  Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

Solution :-
  Here, an = 301, a = 5, d = 11 - 5 = 6
  Now, a + (n-1)d = an
  5 + (n-1) x 6 = 301
  5 + 6n - 6 = 301
  6n - 1 = 301
  6n = 301 + 1
  6n = 302
  n = 302/6
  n = 151/3
  But n should be a positive integer. 
  So, 301 is not a term of the given list of numbers.
    

Example-7 :-  How many two-digit numbers are divisible by 3?

Solution :-
   The list of two-digit numbers divisible by 3 is :
   12, 15, 18, 21, ....... 99
   a = 12, d = 15 - 12 = 3, l = 99
   Now, a + (n-1)d = l
   12 + (n-1) x 3 = 99
   12 + 3n - 3 = 99
   3n + 9 = 99
   3n = 99 - 9
   3n = 90
   n = 30
   So, there are 30 two-digit numbers divisible by 3.
    

Example-8 :-  Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62.

Solution :-
  Here, a = 10, d = 7 - 10 = -3, l = -62
  Now, a + (n-1)d = l
  10 + (n-1) x (-3) = -62
  10 - 3n + 3 = -62
  13 - 3n = -62
  -3n = -62 - 13
  -3n = -75
  n = 75/3
  n = 25
  So, there are 25 terms in the given AP. 
  The 11th term from the last term will be the 15th term.
  So, a15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32 
  i.e., the 11th term from the last term is – 32. 
    
  Alternative Solution : 
  Here, a = -62, d = 3, n = 11
  Now, a11 = a + (n-1)d
  a11 = -62 + (11-1) x (3)
  a11 = -62 + 30
  a11 = -32
  i.e., the 11th term from the last term is – 32. 
    

Example-9 :-  A sum of ₹ 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact.

Solution :-
  Here, Principal (P) = ₹1000, 
         Rate of Interest (R) = 8% Per Year
         Time (T) = 1, 2, 3, .... 30 Years
  Simple Interest (SI) = (P x R x T)/100
  So, Interest at the end of the 1st Year = ₹ (1000 x 8 x 1)/100 = ₹ 80
      Interest at the end of the 2nd Year = ₹ (1000 x 8 x 2)/100 = ₹ 160
      Interest at the end of the 3rd Year = ₹ (1000 x 8 x 3)/100 = ₹ 240
  Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on. 
  So, the interest (in ₹) at the end of the 1st, 2nd, 3rd, . . . years, respectively are 80, 160, 240, . . .
  
  AP : 80, 160, 240, . . .
  a = 80, d = 160 - 80 = 80, n = 30
  Now, an = a + (n-1)d
  a30 = 80 + (30-1) x 80
  a30 = 80 + 2320
  a30 = 2400
  So, the interest at the end of 30 years will be ₹ 2400.
    

Example-10 :-  In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third, and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?

Solution :-
  The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are : 
  23, 21, 19, ......, 5
  It is form in AP. 
  a = 23, d = 21 - 23 = -2, l = 5
  a + (n-1)d = l
  23 + (n-1) x (-2) = 5
  23 - 2n + 2 = 5
  25 - 2n = 5
  -2n = 5 - 25
  -2n = -20
  n = 20/2
  n = 10
  So, there are 10 rows in the flower bed.
    

Example-11 :-  Find the sum of the first 22 terms of the AP : 8, 3, –2, . . .

Solution :-
  Here, a = 8, d = 3 - 8 = -5, n = 22
  Now, Sn = n/2 [2a + (n-1)d]
  S22 = 22/2 [2 x 8 + (22-1) x (-5)]
  S22 = 11 x [16 - 105]
  S22 = 11 x (-89)
  S22 = -979
  So, the sum of the first 22 terms of the AP is – 979.
    

Example-12 :-  If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution :-
  Here, S14 = 1050, n = 14, a = 10, d = ?
  Now, n/2 [2a + (n-1)d] = Sn
  14/2 [2 x 10 + (14-1) x d] = 1050
  7 x [20 + 13d] = 1050
  140 + 91d = 1050
  91d = 1050 - 140
  91d = 910
  d = 910/91
  d = 10
  Therefore, a20 = 10 + (20-1) x 10 = 10 + 190 = 200
    

Example-13 :-  How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78?

Solution :-
  Here, a = 24, d = 21 - 24 = -3, Sn = 78
  Now, n/2 [2a + (n-1)d] = Sn
  n/2 [2 x 24 + (n-1) x (-3)] = 78
  n[48 - 3n + 3] = 78 x 2
  n[-3n + 51] = 156
  -3n2 + 51n - 156 = 0   
  Taking out common -3
  (-3)(n2 - 17n + 52) = 0
  n2 - 17n + 52 = 0
  n2 - 13n - 4n + 52 = 0
  n(n - 13) - 4(n - 13) = 0
  (n - 13)(n - 4) = 0
  Therefore, n = 13, n = 4
  Both values of n are admissible. 
  So, the number of terms is either 4 or 13. 
    

Example-14 :-  Find the sum of : (i) the first 1000 positive integers, (ii) the first n positive integers

Solution :-
(i) the first 1000 positive integers
  Let S = 1 + 2 + 3 + . . . + 1000
  a = 1, d = 2 - 1 = 1, l = 1000
  Now, Sn = n/2 [a + l]
  Sn = 1000/2 [1 + 1000] = 500 x 1001 = 500500
  So, the sum of the first 1000 positive integers is 500500. 
    
(ii) the first n positive integers 
  Let S = 1 + 2 + 3 + . . . + n
  a = 1, d = 2 - 1 = 1, l = n
  Now, Sn = n/2 [a + l]
  Sn = n/2 [1 + n] = n(n+1)/2
  So, the sum of the first n positive integers is n(n+1)/2. 
    

Example-15 :-  Find the sum of first 24 terms of the list of numbers whose nth term is given by an = 3 + 2n

Solution :-
  Given that : an = 3 + 2n
  a1 = 3 + 2 x 1 = 5
  a2 = 3 + 2 x 2 = 7
  a3 = 3 + 2 x 3 = 9
  Here, List of numbers becomes 5, 7, 9, 11, . . .
  a = 5, d = 7 - 5 = 2, n = 24
  Now, Sn = n/2 [2a + (n-1)d]
  S24 = 24/2 x [2 x 5 + (24 - 1) x 2]
  S24 = 12 x [10 + 46]
  S24 = 12 x 56
  S24 = 672
  So, sum of first 24 terms of the list of numbers is 672. 
    

Example-16 :-  A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the 1st year
(ii) the production in the 10th year
(iii) the total production in first 7 years

Solution :-
(i) Since the production increases uniformly by a fixed number every year, 
  the number of TV sets manufactured in 1st, 2nd, 3rd, . . ., years will form an AP. 
  Let us denote the number of TV sets manufactured in the nth year by an. 
  Then, a3 = a + 2d = 600 .....(1)  
      a7 = a + 6d = 700 .....(2) 
  By Solving equation (1) & (2)
  a + 2d - a - 6d = 600 - 700
  -4d = -100
  4d = 100
  d = 100/4
  d = 25
  Putting d = 25 in equation (1)
  a + 2 x 25 = 600
  a + 50 = 600
  a = 600 - 50
  a = 550
  Therefore, production of TV sets in the first year is 550. 
    
(ii)  Now a10 = a + 9d = 550 + 9 × 25 = 775 
  So, production of TV sets in the 10th year is 775. 
    
(iii) S7 = 7/2 x [2 x 550 + (7-1) x 25]
  S7 = 7/2 x [1100 + 150]
  S7 = 7/2 x 1250
  S7 = 7 x 625
  S7 = 4375
  Thus, the total production of TV sets in first 7 years is 4375.
    
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