## 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 15th 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 15th year year we have to find the 15th term of the A.P

tn  =  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 25th month savings

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

a  =  640   d  =  t2 – t1

d  =  720- 640

d  =  80

tn = 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 20th term

tn = 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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