MHB Finding terms in arithmetic progressions

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A company is distributing $36,000 in bonuses to its top five salespeople, with the fifth receiving $6,000 and a constant difference between bonuses. To find the bonuses, the arithmetic sequence formula is applied, leading to the first salesperson receiving $12,000, the second $9,000, and the third $6,000. Additionally, the third term of a recursively defined sequence starting at 3, where each term is five times the previous term minus 2, is calculated to be 63. The sequence progresses as follows: a_1 = 3, a_2 = 13, a_3 = 63. This demonstrates the application of arithmetic sequences and recursive definitions in problem-solving.
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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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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 $$S_n = \dfrac{n}{2}(2a+(n-1)d)$$ where $$n$$ is the number of terms, $$a$$ is the first term and $$d$$ is the common difference.

You're given values for $$S_n$$, $$a$$ and $$n$$ in the question
 
Hello, doreent0722!

4) Find the third term of the recursively defined infinite sequence.
. . a_1=3,\;\;a_{k+1} = 5a_k -2
Do you understand what you are given?

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

. . \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}
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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