# WORD PROBLEMS INVOLVING ARITHMETIC SERIES

Word Problems Involving Arithmetic Series :

In this section, we will learn, how we solve word problems involving arithmetic series.

How to find the find the sum of arithmetic series ?

We use the formulas given below to find the sum of an arithmetic series.

Sn  =  (n/2) [a + l]

Sn  =  (n/2)[2a + (n - 1)d]

To find the number of terms, we use the formula

n  =  [(l - a)/d] + 1

Here a = first term, d = common difference, n = number of terms  and  l  = last term.

## Word Problems Involving Arithmetic Series - Examples

Question 1 :

Find the sum of odd numbers between 0 and 50.

Solution :

Odd numbers between 0 and 50 are

1, 3, 5, 7, ........49.

Since we find the sum of odd numbers lies between 0 and 50, we write the sequence as series.

1 + 3 + 5 + 7 + .............. + 49

a  =  1,  d  =  3 - 1  =  2, l  =  49

To find the total number of terms of the series, we use the formula

n  =  [(l - a)/d] + 1

n  =  [(49 - 1)/2] + 1

n  =  24 + 1  =  25

Sum of 25 terms  :

S=  (n/2) [2 a + (n - 1) d]

S25  =  (25/2) [ 2(1) + (25 - 1) (2)]

=  (25/2) [ 2 + 24 (2)]

=  (25/2) [ 50]

=  625

Hence, the sum of odd numbers between 0 and 50 is 625.

Question 2 :

A contract on construction job specifies a penalty for delay for completion beyond a certain due date as follows. \$ 200 for the first day, \$ 250 for the second day, \$ 300 for the third day etc., the penalty for each succeeding day being \$ 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work be 30 days.

Solution :

For the first day of delay, he has to pay \$200

for the second day, he has to pay \$250

By writing the penalty amount as sequence, we get

200, 250, 300,.............

a = 200,  d = 250 - 200  =  50,  n = 30

To find the money that has to be paid for 30 days of delay, we have to find the sum of 30 terms. So, we are rewriting the sequence as series.

200 + 250 + 300 + ...........

Sn  =  (n/2) [2 a + (n - 1) d]

S30  =  (30/2) [2(200) + (30-1) (50)]

=  15 [400 + 29(50)]

=  15 [400 + 1450]

=  15 [1850]

=  27750

Hence, the total amount for penalty is \$27750.

Question 3 :

A sum of \$700 is to be used to seven cash prizes to students of a school for their overall academic performance. If each prize is \$ 20 less than the preceding prize, find the value of each prizes.

Solution :

The total amount to be spent for prize  =  700

Let "x" be the money value of 1st prize.

Second prize  =  x - 20, third prize  =  x - 40 and so on.

x + (x - 20) + (x - 40) + ...............  =  700

Sn  =  700

a  =  x, d  =  x - 20 - x  =  -20 and n = 7

Sn  =  (n/2) [ 2 a + (n - 1) d]

S7  =  (7/2) [2(x) + (7-1) (-20)]

700  =  (7/2) [2x + 6(-20)]

700  =  (7/2) [2x - 120]

200  =  2x - 120

2x  =  200 + 120

2x  =  320

x  =  320/2  =  160

Hence the money value of prizes are 160, 120, 100, 80, 60, 40 and 20.

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

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