Finding terms in arithmetic progressions

In summary, the company will distribute \$36,000 in bonuses to its top five sales people, with the fifth salesperson receiving \$6,000 and the difference in bonus money between successively ranked salespeople being constant. To find the bonus for each salesperson, we can use the formula S_n = \dfrac{n}{2}(2a+(n-1)d), where n is the number of terms, a is the first term, and d is the common difference. For the recursively defined infinite sequence, the third term can be found by multiplying the previous term by 5 and subtracting 2, starting with a_1 = 3.
  • #1
doreent0722
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3). A company is to distribute \$36,000 in bonuses to its top five sales people. The fifth salesperson on the list will receive \$6,000 and the difference in bonus money between successively ranked salespeople is to be constant. find the bonus for each salesperson.

4). Find the third term of the recursively defined infinite sequence.
[a][/1]=3 [a][k+1]=[5][/ak]-2
 
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  • #2
doreent0722 said:
3). A company is to distribute \$36,000 in bonuses to its top five sales people. The fifth salesperson on the list will receive \$6,000 and the difference in bonus money between successively ranked salespeople is to be constant. find the bonus for each salesperson.

4). Find the third term of the recursively defined infinite sequence.
[a][/1]=3 [a][k+1]=[5][/ak]-2

The sum of an arithmetic sequence is given by \(\displaystyle S_n = \dfrac{n}{2}(2a+(n-1)d)\) where \(\displaystyle n\) is the number of terms, \(\displaystyle a\) is the first term and \(\displaystyle d\) is the common difference.

You're given values for \(\displaystyle S_n\), \(\displaystyle a\) and \(\displaystyle n\) in the question
 
  • #3
Hello, doreent0722!

4) Find the third term of the recursively defined infinite sequence.
. . [tex]a_1=3,\;\;a_{k+1} = 5a_k -2[/tex]

Do you understand what you are given?

The first term is 3.
Thereafter, each term is 5 times the preceding term, minus 2.

. . [tex]\begin{array}{ccccccc} a_1 &=& 3 \\ a_2 &=& 5(3)-2 &=& 13 \\ a_3&=& 5(13) - 2 &=& 63 & {\color{red}\Longleftarrow} \\ a_4 &=& 5(63)-2&=& 313 \\ a_5 &=& 5(313)-2 &=& 1563 \end{array}[/tex]
 

FAQ: Finding terms in arithmetic progressions

What is an arithmetic progression?

An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. For example, 2, 4, 6, 8, 10 is an arithmetic progression with a common difference of 2.

How do you find the common difference in an arithmetic progression?

To find the common difference in an arithmetic progression, you can subtract two consecutive terms. The result will be the constant difference between all terms in the sequence.

What is the formula for finding the nth term in an arithmetic progression?

The formula for finding the nth term in an arithmetic progression is:
an = a1 + (n-1)d
Where an is the nth term, a1 is the first term, and d is the common difference.

Can you find a term in an arithmetic progression without knowing the common difference?

Yes, you can find a term in an arithmetic progression without knowing the common difference. However, you will need to know at least two other terms in the sequence in order to use the formula an = a1 + (n-1)d. If you only know one term, you can use the formula an = 2an-1 - an-2 to find the nth term.

How can arithmetic progressions be applied in real life?

Arithmetic progressions can be applied in many real-life situations, such as calculating compound interest, predicting population growth, and analyzing stock market trends. They are also commonly used in mathematics and computer science to solve problems and create efficient algorithms.

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