# Applications of Series

Arithmetic and geometric series are used in solving real life problems involving patterns. Choose the correct type of series to find a term or a sum of terms.

1. A stack of cans on a display at a supermarket has 6 cans on the top row. The next row down has 2 more cans and the next one has 2 more cans and so on.

a) Calculate the number of cans in the 8th row down.

b) If there are 99 cans in the display altogether, how many rows are there?

The first row has 6 cans, the 2nd row has 9 cans, the 3rd row 12 cans and so on.

a = 6, d = 3

a) For the 8th row, we want n = 8.

${ T }_{ n }=a+\left( n-1 \right) d\\ \\ { T }_{ 8 }=6+\left( 8-1 \right) 2\\ \\ { T }_{ 8 }=6+(7)2\\ \\ { T }_{ 8 }=6+14=20$

There are 20 cans in the 8th row.

b) If there are 450 cans altogether, this is the sum of cans in all rows.

So,

${ S }_{ n }=450\\ \\ { S }_{ n }=\cfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \\ \\ 450=\cfrac { n }{ 2 } \left[ 2\times 6+\left( n-1 \right) 2 \right] \\ \\ 450=\cfrac { n }{ 2 } \left[ 12+2n-2 \right] \\ \\ 450=\cfrac { n }{ 2 } \left[ 2n+10 \right] \\ \\ 450={ n }^{ 2 }+5n\\ \\ { n }^{ 2 }+5n-450=0\\ \\ (n+25)(n-20)=0\\ \\ n=-25\quad OR\quad n=20$

Since n must be a positive integer, then n = 20.

There are 20 rows of cans.

2. A ball is dropped from a height of 1 metre and bounces up to a third of its height. It continues bouncing, rising a third of its previous height on each bounce until it reaches the ground. What is the total distance through which it travels?

We understand that the distance keeps going down and thus the r < 1. The common ratio is 1/3, which leads this geometric series to have a limiting sum.

The total distance can be calculated as follows:

$1+\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 27 } +\cfrac { 1 }{ 27 } +...\\ \\ =1+2\left( \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 27 } +... \right) \\ \\ a=\cfrac { 1 }{ 3 } ,\quad r=\cfrac { 1 }{ 3 } \\ \\ { S }_{ \infty }=\cfrac { a }{ 1-r } \\ \\ { S }_{ \infty }=\cfrac { \cfrac { 1 }{ 3 } }{ 1-\cfrac { 1 }{ 3 } } \\ \\ { S }_{ \infty }=\cfrac { \cfrac { 1 }{ 3 } }{ \cfrac { 2 }{ 3 } } \\ \\ { S }_{ \infty }=\cfrac { 1 }{ 3 } \times \cfrac { 3 }{ 2 } \\ \\ { S }_{ \infty }=\cfrac { 1 }{ 2 } \\ \\ Total\quad distance=1+2\left( \cfrac { 1 }{ 2 } \right) \\ \\ 1+1=2$

So the ball travels 2 metres altogether.

## Compound Interest

In general the compound interest formula is:

$A=P{ (1+r) }^{ n }$

where = principal, r = interest rate, n = number of time periods

### Examples

1. Find the amount that will be in the bank after 6 years if $2000 is invested at 12% p.a. with interest paid: a) yearly b) quarterly a) $P=2000\\ r=12 \% =0.12 \\ n=6 \\ \\ A=2000{ \left( 1+0.12 \right) }^{ 6 }\\ A=2000{ \left( 1.12 \right) }^{ 6 }\\ A=3947.65$ So, the final amount is$3,947.65.

b) For quarterly interest, it is divided into 4 amounts each year.

$r=12\%\div 4=3\%\\ r=0.03\\ \\ n=6\times 4=24\quad quarters\\ \\ A=2000{ \left( 1+0.03 \right) }^{ 24 }\\ A=2000{ \left( 1.03 \right) }^{ 24 }\\ A=4065.59$

So, the final amount is $4,065.59. 2. An amount of$1800 was invested at 6% p.a. and is now worth $2722.66. For how many years was the money invested if interest is paid twice a year? P = 1800, A = 2722.66 Interest is paid twice a year $r=6 \%=3\%\\ r=0.03\\ \\ 2722.66=1800{ \left( 1+0.03 \right) }^{ n }\\ 2722.66=1800{ \left( 1.03 \right) }^{ n }\\ \\ \cfrac { 2722.66 }{ 1800 } ={ \left( 1.03 \right) }^{ n }\\ \\ 1.5126={ 1.03 }^{ n }\\ \log { 1.5126 } =\log { { 1.03 }^{ n } } \\ \log { 1.5126 } =n\log { 1.03 } \\ \\ n=\cfrac { \log { 1.5126 } }{ \log { 1.03 } } \\ \\ n=14$ Since interest is paid in twice a year, the number of years will be 14 / 2 = 7 years. ## Annuities An annuity is a fund where a certain amount of money is invested regularly (often annually) for a period of time. While the interest rate changes, ideally we use a constant amount of interest annually. How do annuities work? Analyse the following example: A sum of$1500 is invested at the beginning of each year in a superannuation fund. If interest is paid at 6% p.a., how much money will be available at the end of 25 years?

We understand that the first amount earns interest for 25 years, the 2nd amount for 24 years, the 3rd amount for 23 years and so on.

$P=1500,\quad r=0.06\\ A=P{ (1+r) }^{ n }\\ { A }_{ 1 }=1500{ \left( 1+0.06 \right) }^{ 25 }\\ { A }_{ 2 }=1500{ \left( 1+0.06 \right) }^{ 24 }\\ { A }_{ 3 }=1500{ \left( 1+0.06 \right) }^{ 23 }\\ .\\ .\\ .\\ { A }_{ 25 }=1500{ \left( 1+0.06 \right) }^{ 1 }\\ \\ Total={ A }_{ 1 }+{ A }_{ 2 }+{ A }_{ 3 }+...+{ A }_{ 25 }\\ \qquad \quad =1500{ \left( 1.06 \right) }^{ 25 }+1500{ \left( 1.06 \right) }^{ 24 }+1500{ \left( 1.06 \right) }^{ 23 }+...+1500{ \left( 1.06 \right) }^{ 1 }\\ \qquad \quad =1500\left( { 1.06 }^{ 25 }+{ 1.06 }^{ 24 }+{ 1.06 }^{ 23 }+...+{ 1.06 }^{ 1 } \right) \\ \\ { 1.06 }^{ 1 }+{ 1.06 }^{ 2 }+{ 1.06 }^{ 3 }+...+{ 1.06 }^{ 25 }\quad is\quad a\quad geometric\quad series\\ a=1.06,\quad r=1.06,\quad n=25\\ \\ { S }_{ n }=\cfrac { a{ \left( { r }^{ n }-1 \right) } }{ r-1 } \\ \\ { S }_{ 25 }=\cfrac { 1.06{ \left( 1.06^{ 25 }-1 \right) } }{ 1.06-1 } \\ \\ { S }_{ 25 }=58.16\\ \\ Total=1500\times 58.16\\ \qquad \quad =87234.57$

Therefore, the superannuation after 25 years is $87 234.57. ## Loan Repayments A sum of$20 000 is borrowed at 12% p.a. and paid back at regular monthly intervals over 4 years. Find the amount of each payment.

Let M stand for the monthly repayment.

The number of payments will be 4 x 12 or 48.

Monthly interest will be 12% / 12 = 1% or 0.01.

The interest added to the first month is $A{(1+0.01)}^{1} or A{1.01}^{1}$. The amount owing for the first month:

${A}_{1}=20000{(1.01)}^{1}-M$

The amount owing for the 2nd month is what was owing from the 1st month, together with that month's interest, minus the repayment.

${ A }_{ 2 }={ A }_{ 1 }{ \left( 1.01 \right) }^{ 1 }-M\\ \quad \quad =[20000{ (1.01) }^{ 1 }-M]{ \left( 1.01 \right) }^{ 1 }-M\\ \quad \quad =20000{ (1.01) }^{ 2 }-M{ (1.01) }^{ 1 }-M\\ \quad \quad =20000{ (1.01) }^{ 2 }-M({ 1.01 }^{ 1 }+1)\\ \\ { A }_{ 3 }={ A }_{ 2 }{ (1.01) }^{ 1 }-M\\ \quad \quad =[20000{ (1.01) }^{ 2 }-M({ 1.01 }^{ 1 }+1)]{ (1.01) }^{ 1 }-M\\ \quad \quad =20000{ (1.01) }^{ 3 }-M({ 1.01 }^{ 1 }+1){ (1.01) }^{ 1 }-M\\ \quad \quad =20000{ (1.01) }^{ 3 }-M{ ({ 1.01 }^{ 2 } }+{ 1.01 }^{ 1 })-M\\ \quad \quad =20000{ (1.01) }^{ 3 }-M({ { 1.01 }^{ 2 } }+{ 1.01 }^{ 1 }+1)\\ \\ Continuing\quad the\quad pattern\quad after\quad 48\quad years:\\ \\ { A }_{ 48 }=20000{ (1.01) }^{ 48 }-M({ { 1.01 }^{ 47 } }+{ 1.01 }^{ 46 }+{ 1.01 }^{ 45 }+...+1)\\ \\ But\quad the\quad loan\quad is\quad paid\quad out\quad after\quad 48\quad months.\quad So,\\ \\ 20000{ (1.01) }^{ 48 }-M({ { 1.01 }^{ 47 } }+{ 1.01 }^{ 46 }+{ 1.01 }^{ 45 }+...+1)=0\\ \\ M=\cfrac { 20000{ (1.01) }^{ 48 } }{ ({ { 1.01 }^{ 47 } }+{ 1.01 }^{ 46 }+{ 1.01 }^{ 45 }+...+1) } \\ \\ 1+{ 1.01 }^{ 1 }+{ 1.01 }^{ 2 }+{ 1.01 }^{ 3 }+...+{ 1.01 }^{ 48 }\\ \\ Geometric\quad series:\\ a=1,\quad r=1.01,\quad n=48\\ \\ { S }_{ n }=\cfrac { a\left( { r }^{ n }-1 \right) }{ r-1 } \\ \\ { S }_{ 48 }=\cfrac { 1\left( { 1.01 }^{ 48 }-1 \right) }{ 1.01-1 } \\ \\ { S }_{ 48 }=61.223\\ \\ \therefore \quad M=\cfrac { 20000{ (1.01) }^{ 48 } }{ 61.223 } =526.68$

So the monthly repayment is \$526.68.