Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
4. Arithmetic Series
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Exercise 14 Page 591

Practice makes perfect
a Let's write an expression for the given arithmetic sequence. We can start by finding the common difference. To do so we will subtract consecutive terms.
common difference
We see that the first term is 75 and that the common difference is 4. Let's substitute these values in the explicit rule of arithmetic sequences.
a_n = a_1 + d(n-1)
a_n = 75 + 4(n-1)
Simplify right-hand side
a_n = 75+4n-4
a_n = 71+4n
We found the formula of our arithmetic sequence. The grade of the student on his sixth test is the sixth term of the arithmetic sequence. To find its value, we will substitute 6 for n in the obtained formula.
a_n = 71 + 4n
a_6 = 71 + 4( 6)
Evaluate right-hand side
a_6= 71+24
a_6 =95
If the student continues to improve at the same rate, the sixth test will have a score of 95.
b Let's use the formula for the sum of a finite arithmetic series to find the sum of our series. First, we can set up our summation for the six tests using the formula we derived from Part A.
S_6= ∑_(n=1)^6 (71+4n) We know the first term is a_1 = 75, the last term is a_6 = 95, and that the series has 6 terms. Let's substitute these values into the sum formula to find the sum of the six terms in the series.
S_n=n/2(a_1+a_n)
S_6= 6/2(a_1+a_6)
S_6=6/2( 75+ 95)
Evaluate right-hand side
S_6 = 3(75+95)
S_6 = 3(170)
S_6 = 510
The total score for the six tests is 510.