Application of Arithmetic Sequence and Series

Example:

Tickets for a certain show were printed bearing numbers from 1 to 100. Odd number tickets were sold by receiving cents equal to thrice of the number on the ticket while even number tickets were issued by receiving cents equal to twice of the number on the ticket. How much amount was received by the issuing agency?

Solution:

Let {S_1} and {S_2} be the amounts received for odd number and even number ticket respectively, then
{S_1} = 3\left[ {1 + 3 + 5 + \cdots + 99} \right] and {S_2} = 2\left[ {2 + 4 + 6 + \cdots + 100} \right]

Thus, {S_1} + {S_2} = 3 \times \frac{{50}}{2}\left( {1 + 99} \right) + 2 \times \frac{{50}}{2}\left( {2 + 100} \right)
\because There are 50 terms in each series
{S_1} + {S_2} = 7500 + 5100 = 12600

Hence the total amount received by the issuing agency  = 12600 cents  = \$ 126

Example:

A clock strikes once when its hour hand is at 1, twice when it is at 2 and so on. How many times does the clock strike in six hours?

Solution:

Since the clock strikes once when its hour hand is at 1, twice when it is at 2 and so on, so the sequence of strikes from 1 hour to 6 hours is 1,2,3,4,5,6.

Here {a_1} = 1, d = 2 - 1 = 1, n = 6 so
{S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]
 = \frac{6}{2}\left[ {2\left( 1 \right) + \left( {6 - 1} \right)\left( 1 \right)} \right] = 3\left( {2 + 5} \right) = 3\left( 7 \right) = 21

This shows that the clock strikes 21times in6hours.

Example:

A factory owner repays his loan of \$ 2088000 by \$ 20000 in the first monthly installment and then increases the payment by \$ 1000 in every installment. In how many installments he will clear his loan?

Solution:

By the given conditions, we have
{a_1} = 20000, d = 1000, {S_n} = 2088000
{S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]
2088000 = \frac{n}{2}\left[ {2\left( {20000} \right) + \left( {n - 1} \right)\left( {1000} \right)} \right]
 \Rightarrow 2088 = \frac{n}{2}\left[ {40 + n - 1} \right]
 \Rightarrow 4176 = {n^2} + 39n
 \Rightarrow {n^2} + 39n - 4176 = 0
 \Rightarrow n = \frac{{ - 39 \pm \sqrt {{{\left( {39} \right)}^2} - 4\left( 1 \right)\left( { - 4176} \right)} }}{{2\left( 1 \right)}} = \frac{{ - 39 \pm \sqrt {1521 + 16704} }}{2}
 = \frac{{ - 39 \pm \sqrt {18225} }}{2} = \frac{{ - 39 \pm 135}}{2}
 = \frac{{ - 39 + 135}}{2},\frac{{ - 39 - 135}}{2}
 \Rightarrow n = 48, - 87

Now n, being the number of installments, cannot be negative, so n = 48. This shows that the factory owner will clear his loan in 48 monthly installments.