PRACTICAL PROBLEMS IN ARITHMETIC SEQUENCE

Practical Problems in Arithmetic Sequence :

An arithmetic series is a series whose terms form an arithmetic sequence.

Problems in Arithmetic Sequence

Problem 1 :

A TV manufacturer has produced 1000 TVs in the seventh year and 1450 TVs in the tenth year. Assuming that the production increases uniformly by a fixed number every year, find the number of TVs produced in the first year, find the number of TVs produced in the first year and 15^th year.

Solution :

Number of TVs produced in the seventh year  =  1000

Number of TVs produced in the tenth year  =  1450

t₇  =  1000

t₁₀  =  1450

a + 6d  =  1000   ----- (1)

a + 9d  =  1450   ----- (2)

(1) – (2)

  Subtracting second equation from first equation

aaaaaaaaaaaaaaa + 6d  =  1000aaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaaa + 9d  =  1450aaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaa(-)    (-)      (-)aaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaa------------------aaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaa-3d = -450aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

aaaaaaaaaaaaaad = -450/(-3) ==> d = 150

By applying d  =  150 in (1)

 a + 6(150)  =  1000

  a + 900  =  1000

 a  =  1000 -900

  a  =  100

Therefore number of TVs produced on the first year is 100

To find number of TVs produced in the 15^th year year we have to find the 15^th term of the A.P

t_n  =  a + (n-1) d

t₁₅  =  100 + (15-1) 150

t₁₅ =  100 + 14(150)

    t₁₅  =  100 + 2100

    t₁₅  =  2200

Problem 2 :

A man has saved $640 during the first month,$720 in the second month and $800 in the third month. If he continues his savings in this sequence, what will be his savings in the 25th month?

Solution :

If we write his saving like a sequence, we will get

640, 720, 800,…………

to get 25^th month savings

From this, we have to find the 25^th term of the sequence

a  =  640   d  =  t₂ – t₁

d  =  720- 640

d  =  80

t_n = a + (n-1) d

t₂₅ = 640 + (25 - 1) 80

  =  640 + 24 (80)

  =  640 + 1920

  =  2560

Hence he saves $2560 in the 25th month

Problem 3 :

A person has deposited $25000 in an investment which yields 14% simple interest annually. Do these amounts (principal + interest) form an A.P ? If so, determine the amount of investment after 20 years.

Solution :

Simple interest = PNR/100

   =  (25000 x 1 x 14)/100

  = 3500

Amount = principal + interest

 = 25000 + 3500

= 28500

Amount at the end of the first year = 28500

Amount at the end of second year = 28500 + 3500

                                                       = 32000

Amount at the end of third year = 32000 + 3500

                                                  = 35500 

28500, 32000, 35500.,………………….

This is the arithmetic sequence. To find the amount of investment after 20 years we need to find 20^th term

t_n = a +(n - 1) d

a = 28500    d = 32000 – 28500

d  =  3500

t₂₀  =  28500 + (20 - 1) 3500

t₂₀  =  28500 + 19 (3500)

      t₂₀  =  28500 + 66500

     t₂₀  =  95000

Therefore the amount of investment in after years is 95000.

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