Problem Study (Arithmetic Progression)

1.    The Length of a perimeter of a hexagon is 36cm. The lengths the sides of the
       hexagon are in arithmetic progression and the length of the longest side is five times
       the length of the shortest side. Find the length of each side.

      Solution 

       Let the length of the sides of the hexagon in ascending order be u₁, u₂, u₃, u₄,
       u₅ and u₆.
       By the problem u₁, u₂, u₃, u₄, u₅, u₆ is an A.P.
       Let u₁ = a,
             u₂ = a + d
             u₃ = a + 2d
             u₄ = a + 3d
             u₅ = a + 4d
             u₆ = a + 5d
      Perimeter of the hexagon = 36 (given)
      Hence, 6a + 15d = 36
                 2a + 5d = 12 ............ (1)
      length of longest side = 5 ( length of shortest side)
      u₆ = 5u_{1
       ∴}  a + 5d = 5a
           4a - 5d = 0 .............(2)
      Solving equation (1) and (2), a = 2 and d = ${\displaystyle \frac{8}{\mathrm{5 }}}$= 1.6
      u₁ = a = 2 cm
      u₂ = a + d = 3.6 cm
      u₃ = a + 2d = 5.2 cm
      u₄ = a + 3d = 6.8 cm
      u₅ = a + 4d = 8.4 cm
      u₆ = a + 5d = 10 cm

2.   The sum of n terms of an arithmetic progression is given by the formula S_n = 2n² + n.
      Find (a) the first term, (b) the common difference and (c) the tenth term.

     Solution 

      S_n = 2n² + n
      u₁ = S₁ = 2(1)² + (1) = 3
      _{∴ the first term = a = 3}
      u₁ + u₂ = S₂ = 2(2)² + (2) = 10
      ∴ u₂ = S₂ - S₁ = 7
      ∴ the common difference = d = u₂ - u₁ = 4
      u_n = a + (n - 1)d
      ∴ u₁₀ = a + 9d= 3 + 9(4) = 39

3.   During 1996 a company increased its sales of television sets at a constant rate of 200
      sets per month. Thus the number of television sets sold in February was 200 more
      than in January, the number of television sets sold in March was 200 more than in
      February and this pattern continued month by month throughout the year. Given that
      the company sold 38, 400 television sets in 1996, calculate the number of television
      sets sold in (i) January,  (ii) December.

     Solution

      Let the number of television sets sold in January = a
      and those sold in December = u₁₂.       
      ∴ d = 200, n = 12 and S₁₂ = 38, 400
      S_n  = ${\displaystyle \frac{n}{2}}$  { 2a + (n - 1) d } 
      S₁₂ = 38, 400
      ∴  ${\displaystyle \frac{\mathrm{12}}{2}}$  (2a + 11 × 200) = 38,400 
      2a + 2,200  = 6,400
      a  =2100
      u_n  = a + (n - 1)d 
      u₁₂  = a + 11d = 2,100 + 11 (200) = 4,300
      Therefore there were 2,100 television sets sold in January and 4,300 sets in December.

4.   Find the sum of all numbers between 200 and 1,000 which are exactly divisible by 15.

     Solution

      All numbers between 200 and 1,000 which are exactly divisible by 15 are
      210, 225, 240, ... , 990.
      Here the terms are in an A.P., with a = 210, d = 15 and l = u_n  = 990.
      Since  u_n  = a + (n - 1)d,  
      a + (n - 1)d = 990
      210 + (n - 1)(15) = 990
      n = 53 
      S_n  = ${\displaystyle \frac{n}{2}}$ (a + l)
      S₅₃  =  ${\displaystyle \frac{53}{2}}$ (210 + 990) = 31,800

Credit : Problems Supported by : Sayar Idea Zaw