PRACTICAL 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

-3d = -450

d = -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.

Problem 4 :

A mother divides $207 into three parts such that the amount are in A.P and gives it to her three children. The product of the two least amounts that the children had $4623. Find the amount received by each child.

Solution :

Let a - d, a and a + d be the amount she is giving to each children.

a - d + a + a + d = 207

3a = 207

a = 207/3

a = 69

The product of two least amount = 4623

a(a - d) = 4623

69(69 - d) = 4623

69 - d = 4623/69

69 - d = 67

d = 69 - 67

d = 2

The amount received by the first child = a - d

= 69 - 2

= 67

The amount received by the first child = a

= 69

The amount received by the first child = a + d

= 69 + 2

= 71

So, the amount received by each child are

$67, $69 and $71.

Problem 5 :

In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?

Solution :

Number of seats in the first row (a) = 20

Difference between the number of seats (d) = 2

Total number of rows (n) = 30

s_n = (n/2) [2a + (n - 1)d]

= (30/2)[2(20) + (30 - 1)2]

= 15[40 + 29(2)]

= 15[40 + 58]

= 15(98)

= 1470

So, the total number of seats is 1470.

Problem 6 :

In a winter season let us take the temperature of Ooty from Monday to Friday to be in A.P. The sum of  temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.

Solution :

a = temperature on Monday

a + d = Temperature on Tuesday

a + 2d = Temperature on Wednesday and so on

Sum of temperatures from Monday to Wednesday = 0

a + a + d + a + 2d = 0

3a + 3d = 0

a + d = 0 ------(1)

Sum of temperatures from Wednesday to Friday = 0

a + 3d + a + 4d + a + 5d = 18

3a + 12d = 18

Dividing by 3, we get

a + 4d = 6 ------(2)

(1) - (2)

d - 4d = 6

-3d = 6

d = -2

Applying the value of d in (1), we get

a - 2 = 0

a = 2

a + d = 0

a + 2d = 2 + 2(-2) ==> -2

a + 3d = 2 + 3(-2) ==> -4

a + 4d = 2 + 4(-2) ==> -6

Temperature on Monday to Friday 

2, 0, -2, -4, -6

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