# Application of Arithmetic Sequence and Series

Example:

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

Solution:

Let ${S_1}$ and ${S_2}$ be the amounts received for odd number and even number tickets 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 $21$ times in $6$ hours.

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. How many installments will it take for him to 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.