Question

1)an arithmetic progression has a first term of 32 , a 5th term of 22 and a last term of -28 .find the sum of all the terms in the progression .

2)each year a school allocates a sum of money for the library . the amount allocated each year increases by 2.5 % of the amount allocated the previous year . in 2005 the school allocated $2000.find the total amount allocated in the years 2005 to 2014 inclusive

Answer 1

1.)

The first term in arithmetic progression is $(a) = 32$

Since the $5^{th}$ term in the arithmetic progression is $22$ so we can write:

$t_5 = 22 \\ \Rightarrow a + (5-1)d = 22$ $\hspace{1 cm} [ \because$ $t_n = a + (n-1)d \, \,$ ( Where a = first term and d = common difference) ]

$\Rightarrow a + 4d = 22 \\ \Rightarrow 32 + 4d = 22 \\ \Rightarrow 4d = 22 - 32 \\$

$\Rightarrow d = - \frac{10}{4} \\ \Rightarrow d = - 2.5$

Let us assume that number of terms in this progression be $n$

Now:

Since the last term of this progression is $-28$ so we can write:

$t_n = -28 \\ \Rightarrow a + (n-1)d = -28 \\ \Rightarrow 32 + (n-1)(-2.5) = -28 \hspace{1 cm}$ [ $\because a = 32 \, \,\, and \,\,\, d = -2.5$ ]

$\Rightarrow n-1 = \frac{ -28 - 32}{-2.5} \\ \Rightarrow n - 1 = 24 \\ \Rightarrow n = 25$

Hence:

The sum of all terms in this progression is:

$S_n = \frac{n}{2} (a + l) \hspace{1 cm}$ [ Where a = first term and l = last term ]

$\hspace{0.5 cm} = \frac{25}{2} (32 -28) \\ \hspace {0.5 cm} = 50$

2.)

Since the amount of money each year increased by $2.5 \%$ of the amount of money allocated in the previous year so each year added a fixed amount money of the money allocated in the previous year and allocated money from $2005 \, \text{to} \, 2014$ forms an arithmetic progression.

Now:

Amount of money allocated in $2005$ is $\$ \, 2000$ which is also a first term (a) in this progression.

Number of terms in this progression is:

$n =$ number of years from $2005 \, \text{to} \, 2014 = 10$

Amount of money allocated in 2006 is:

$= \$ \, \left( 2000 + 2000 \times 2.5 \% \right) \\ = \$ \, 2050$

So, common difference in this progression is:

$d = 2050 - 2000 = 50$ [ $\because$ Common difference in an arithmetic progression (d) = next term - previous term ]

Hence:

The total amount of money allocated from $2005 \, \text{to} \, 2014$ inclusive is:

$= \frac{n}{2} \left( 2a + (n-1)d \right) \\$

$= \frac{10}{2} \left( 2(2000) + (10-1)50 \right) \\ = \$ \, 22250$