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lara31 [8.8K]
3 years ago
13

If 2k, 5k-1 and 6k+2 are the first 3 terms of an arithmetic sequence, find k and the 8th term​

Mathematics
1 answer:
Klio2033 [76]3 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

The common difference d of an arithmetic sequence is

d = a_{2} - a_{1} = a_{3} - a_{2}

Substitute in values and solve for k, that is

5k - 1 - 2k = 6k + 2 - (5k - 1)

3k - 1 = 6k + 2 - 5k + 1

3k - 1 = k + 3 ( subtract k from both sides )

2k - 1 = 3 ( add 1 to both sides )

2k = 4 ⇒ k = 2

--------------------------------------------------------

The n th term of an arithmetic sequence is

a_{n} = a_{1} + (n - 1)d

a_{1} = 2k = 2 × 2 = 4 and

d = 5k - 1 - 2k = 3k - 1 = (3 × 2) - 1 = 5

Hence

a_{8} = 4 + (7 × 5) = 4 + 35 = 39

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Using the binomial distribution, it is found that there is a 0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

For each pill, there are only two possible outcomes, either it is acceptable, or it is not. The probability of a pill being acceptable is independent of any other pill, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

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In this problem:

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The probability that <u>fewer that 5 in a sample of 20 pills</u> will be acceptable is:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0

P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0

P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0

P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0

P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377

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