## REAL LIFE PROBLEMS INVOLVING ARITHMETIC SERIES

real life problems involving arithmetic series :

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

Here we are going to see some practical problems based on the topic arithmetic series.

## Real life problems involving arithmetic series - Examples

Problem 1 :

A construction company will be penalized each day of delay in construction for bridge. The penalty will be \$4000 for the first day and will increase by \$10000 for each following day. Based on its budget, the company can afford to pay a maximum of \$ 165000 toward penalty. Find the maximum number of days by which the completion of work can be delayed

Solution :

Let us write the penalty amount paid by the construction company from the first day as sequence

4000,5000,6000,..............

The company can pay 165000 as penalty for this delay at maximum.

So, we have to write this amount as series 4000 + 5000 + 6000 + ..... and the sum of the penalty amount is 165000.

Sn= 165000

a = 4000 d = 1000

(n/2)[2a+(n-1)d] = 165000

(n/2)[ 2 (4000) + (n-1) 1000 ] = 165000

(n/2)[8000 + (n-1) 1000] = 165000

(n/2)[8000 + 1000 n- 1000] = 165000

(n/2)[7000 + 1000 n] = 165000

n[7000 + 1000 n] = 165000 x 2

7000 n + 1000 n² = 330000

divided by 10000 => 7n + n² - 330

n² + 7 n - 330 = 0

(n-15) (n+22)= 0

n = 15, -22

Here n represents number of days delayed. So it must be positive.Therefore the correct answer is 15.

Problem 2 :

The sum of \$1000 is deposited every year at 8% simple interest. Calculate the interest at the end of each year. Do these interest amounts form an A.P?. If so,find the total interest at the end of 30 years.

Solution :

First let us find the interest amount from simple interest formula.

S.I = (PNR)/100

= (1000 x 1 x 8)/100

= 80

1000,1080,1160,.........

the interest amounts from the first year are 80,160,240..........

This sequence is A.P. From this we have to find the sum of 30 terms . Because we need to find the interest amount for 30 years.

a = 80 d = 160 - 80    n = 30

= 80

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

= (30/2) [2(80) + (30-1) 80]

= 15 [160 + 29(80)]

= 15 [160 + 2320]

= 15 [2480]

= 37200

Problem 3 :

If a clock strikes once at 1'o clock,twice at 2'o clock and so on.How many times will it strike a day?

Solution :

The clock strikes once at 1'o clock,twice at 2'o clock and so on.Now we have to write this pattern as series because we need to find number of times that the clock strike a day.

= 2 [1 + 2 + 3 + ............ + 12]

= 2 x (n/2) [a+L]

= 2 x (12/2) [1+12]

= 12 [13]

= 156

Therefore the clock will strike 156 times in a day.

Problem 4 :

A gardener plans to construct a trapezoidal shaped structure in his garden. The longer side of trapezoid needs to start with a row of 97 bricks. Each row must be decreased by 2 bricks on each end  and the construction should stop at 25th row. How many bricks does he need to buy?

Solution :

The longer side of trapezoid shaped garden is containing 97 and each row must be decreased by 2 on each end and the construction must stop when it reaches the 25th row.

If we write the number of bricks in each row as sequence we will get 97, 93, 89,............

Now we nee to find number of bricks needed to buy. For that we have to make it as series and we have to find for 25 terms of the series.

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

a = 97  d = 93-97   n = 25

= -4

S25 = (25/2) [2(97) + (25-1) (-4)]

= (25/2) [194 + (24) (-4)]

= (25/2) [194 - 96]

= (25/2) (98)

= 25 x 49

= 1225 bricks

After having gone through the stuff given above, we hope that the students would have understood "Real life problems involving arithmetic series".

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