Arithmetic Progressions Examples and Solutions

Ex. (1) A mixer manufacturing company manufactured 600 mixers in 3rd year and in 7th year they manufactured 700 mixers. If every year there is same growth in the production of mixers then find (i) Production in the first year (ii) Production in 10th year (iii) Total production in first seven years.

Solution : Addition in the number of mixers manufactured by the company per year is constant therefore the number of production in successive years is in A.P.
(i) Let’s assume that company manufactured tn mixers in the nth year then as per
    given information,
    t₃ = 600, t₇ = 700
   We know that t_n= a+(n-1)d
      t₃= a+(3-1)d
      a + 2d = 600. . . (I)
      t₇= a+(7-1)d
      t₇= a+6d = 700
      a+2d = 600

     ∴ Substituting a = 600 - 2d in equation (II),
     600 - 2d + 6d = 700
    4d = 100 

    ∴ d = 25
    a+2d = 600

    ∴ a + 2 ×25 = 600
    a + 50 = 600

    ∴ a = 550
    ∴ Production in first year was 550.

(ii) t_n= a+(n-1)d
     t₁₀= 550+(10-1) × 25
         = 550 + 225
     Production in 10th year was 775.

(iii) For finding total production in first 7 years let’s use formula for Sn.
     `S_n =n/2 [2a+(n-1)d]`
     `S_n =7/2 [1100 + 150] =7/2 [1250] = 7 xx 625 = 4375`
     Total production in first 7 years is 4375 mixers.