Answer to Question #110565 in Algebra for atta

Question #110565
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
1
Expert's answer
2020-05-26T20:15:54-0400

1.)

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

Since the 5th5^{th} term in the arithmetic progression is 2222 so we can write:

t5=22a+(51)d=22t_5 = 22 \\ \Rightarrow a + (5-1)d = 22 [\hspace{1 cm} [ \because tn=a+(n1)d  t_n = a + (n-1)d \, \, ( Where a = first term and d = common difference) ]

a+4d=2232+4d=224d=2232\Rightarrow a + 4d = 22 \\ \Rightarrow 32 + 4d = 22 \\ \Rightarrow 4d = 22 - 32 \\

d=104d=2.5\Rightarrow d = - \frac{10}{4} \\ \Rightarrow d = - 2.5

Let us assume that number of terms in this progression be nn

Now:

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

tn=28a+(n1)d=2832+(n1)(2.5)=28t_n = -28 \\ \Rightarrow a + (n-1)d = -28 \\ \Rightarrow 32 + (n-1)(-2.5) = -28 \hspace{1 cm} [ a=32   and   d=2.5\because a = 32 \, \,\, and \,\,\, d = -2.5 ]

n1=28322.5n1=24n=25\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:

Sn=n2(a+l)S_n = \frac{n}{2} (a + l) \hspace{1 cm} [ Where a = first term and l = last term ]

=252(3228)=50\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%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 2005to20142005 \, \text{to} \, 2014 forms an arithmetic progression.

Now:

Amount of money allocated in 20052005 is $2000\$ \, 2000 which is also a first term (a) in this progression.

Number of terms in this progression is:

n=n = number of years from 2005to2014=102005 \, \text{to} \, 2014 = 10

Amount of money allocated in 2006 is:

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

So, common difference in this progression is:

d=20502000=50d = 2050 - 2000 = 50 [ \because Common difference in an arithmetic progression (d) = next term - previous term ]

Hence:

The total amount of money allocated from 2005to20142005 \, \text{to} \, 2014 inclusive is:

=n2(2a+(n1)d)= \frac{n}{2} \left( 2a + (n-1)d \right) \\

=102(2(2000)+(101)50)=$22250= \frac{10}{2} \left( 2(2000) + (10-1)50 \right) \\ = \$ \, 22250


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