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9. -13, -11, -9, -7, $\boldsymbol{-5}$, $\boldsymbol{-3}$, $\boldsymbol…

Question

  1. -13, -11, -9, -7, $\boldsymbol{-5}$, $\boldsymbol{-3}$, $\boldsymbol{-1}$10. -2, -5, -8, -11, $\boldsymbol{-14}$, $\boldsymbol{-17}$, $\boldsymbol{-20}$write an equation to find the $n^{th}$ term of each sequence. then find $a_{24}$11. 1, 3, 5, 7, ...12. -1, -4, -7, -10, ...13. -4, -9, -14, -19, ...14. 7, 13, 19, 25, ...15. charlie deposited $115 in a savings account. each week thereafter, he deposits $35 into the account.a. write a formula to represent this sequence.b. how much total money has charlie deposited after 30 weeks?16. as manager of the soccer team, wendy is to hand out cups of water at practice. each cup of water is 4 ounces. she begins practice with a 128-ounce cooler of water.a. write a formula to represent this sequence.b. how much water is remaining after she hands out the $14^{th}$ cup?© gina wilson (all things algebra® llc), 2012-2018

Explanation:

Sequence 11: 1, 3, 5, 7, ...

Step1: Identify arithmetic sequence parts

First term $a_1=1$, common difference $d=3-1=2$.

Step2: Apply nth term formula

Arithmetic sequence formula: $a_n = a_1 + (n-1)d$
$a_n = 1 + (n-1) \times 2 = 2n - 1$

Step3: Find 42nd term

Substitute $n=42$:
$a_{42} = 2(42) - 1$

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Sequence 12: -1, -4, -7, -10, ...

Step1: Identify arithmetic sequence parts

First term $a_1=-1$, common difference $d=-4-(-1)=-3$.

Step2: Apply nth term formula

$a_n = -1 + (n-1)(-3) = -3n + 2$

Step3: Find 42nd term

Substitute $n=42$:
$a_{42} = -3(42) + 2$

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Sequence 13: -4, -9, -14, -19, ...

Step1: Identify arithmetic sequence parts

First term $a_1=-4$, common difference $d=-9-(-4)=-5$.

Step2: Apply nth term formula

$a_n = -4 + (n-1)(-5) = -5n + 1$

Step3: Find 42nd term

Substitute $n=42$:
$a_{42} = -5(42) + 1$

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Sequence 14: 7, 13, 19, 25, ...

Step1: Identify arithmetic sequence parts

First term $a_1=7$, common difference $d=13-7=6$.

Step2: Apply nth term formula

$a_n = 7 + (n-1) \times 6 = 6n + 1$

Step3: Find 42nd term

Substitute $n=42$:
$a_{42} = 6(42) + 1$

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Problem 15: Charlie's Savings

Part a: Write the formula

Step1: Identify sequence parts

Initial deposit (week 0) $a_1=115$, weekly deposit $d=35$. This is an arithmetic sequence where $n$ = number of weeks.

Step2: Apply savings formula

Total deposit $a_n = 115 + 35(n-1) = 35n + 80$

Part b: Total after 30 weeks

Step1: Substitute n=30 into formula

$a_{30} = 35(30) + 80$

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Problem 16: Wendy's Water Cooler

Part a: Write the formula

Step1: Identify sequence parts

Initial water $a_1=128$, water per cup $d=-4$. Remaining water after $n$ cups:
$a_n = 128 - 4(n-1) = 132 - 4n$

Part b: Water after 14 cups

Step1: Substitute n=14 into formula

$a_{14} = 132 - 4(14)$

Answer:

  1. nth term: $a_n=2n-1$; 42nd term: $83$
  2. nth term: $a_n=-3n+2$; 42nd term: $-124$
  3. nth term: $a_n=-5n+1$; 42nd term: $-209$
  4. nth term: $a_n=6n+1$; 42nd term: $253$
  5. a. $a_n=35n+80$; b. $\$1130$
  6. a. $a_n=132-4n$; b. $76$ ounces