**Practical Problems in Arithmetic Sequence :**

An arithmetic series is a series whose terms form an 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

t_{7} = 1000

t_{10} = 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

t_{15} = 100 + (15-1) 150

t_{15} = 100 + 14(150)

t_{15} = 100 + 2100

t_{15} = 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 = t_{2} – t_{1}

d = 720- 640

d = 80

t_{n} = a + (n-1) d

t_{25} = 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

t_{20} = 28500 + (20 - 1) 3500

t_{20 }= 28500 + 19 (3500)

t_{20 }= 28500 + 66500

t_{20 } = 95000

Therefore the amount of investment in after years is 95000.

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