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fiasKO [112]
3 years ago
10

Which term of the arthemetic sequence (1, 3, 5, 7) is equal to 141

Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
6 0

Answer:

71th term is 141

Step-by-step explanation:

a_{1}=1\\\\common \ difference = d = second \ term - first \ term\\\\ = 3 - 1 = 2\\\\a_{n} = 141\\\\a + (n -1)*d = a_{n}

1 + (n - 1)*2 = 141         {Use distributive property}

1 + 2n - 2 = 141    

2n - 1 = 141

2n = 141 + 1

2n = 142

n = 142/2

n = 71

gregori [183]3 years ago
3 0

Answer:

71st term

Step-by-step explanation:

First, find the nth term: 2n - 1 (the difference is 2, and the 0th term or the number before the 1 is -1). Next, set the nth term equal to 141 and solve for n. 2n - 1 = 141. We would add one to both sides → 2n = 142, then divide both sides by 2. n = 71.

Hope this helps!

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