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stira [4]
2 years ago
12

A sequence can be generated by using an = an-1 + 4, where a1 = 6 and n is a whole number greater than 1. What are the first four

terms in the sequence?
Mathematics
1 answer:
Firlakuza [10]2 years ago
7 0

<u>Answer:</u>

  • The first four terms are <u>4</u><u>,</u><u>8</u><u>,</u><u>1</u><u>2</u><u>,</u><u>1</u><u>6</u>

<u>Step-by-Step </u><u>Explanation</u><u>:</u>

The given relation between the nth term and it's previous term is given by:

\boxed{an = an - 1 + 4}

GiveN:

  • a1 = 4

Now finding the other three terms of the AP with the given relation.

a2 = a1 + 4

Putting a1 = 4,

a2 =  4 + 4 = 8

Now, Third term:

a3 = a2 + 4

Putting a2 = 8,

a3 = 8 + 4 = 12

Now, Fourth term:

a4 = a3 + 4

Putting a3 = 12,

a4 = 12 + 4 = 16

Hence, The first four terms of the AP is 4, 8, 12 & 16.

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Start by subtracting 10 from both sides.

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Step-by-step explanation:

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So, the possibilities that at-least 3 out of 5 students receive version A are,

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2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}+\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}

= (\frac{1}{4})^3\times (\frac{3}{4})^2+(\frac{1}{4})^4\times (\frac{3}{4})+(\frac{1}{4})^5

= (\frac{1}{4})^3\times (\frac{3}{4})[\frac{3}{4}+\frac{1}{4}+(\frac{1}{4})^2]

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= 0.01171875 × 1.0625

= 0.01245

Thus, the probability that at least 3 out of 5 students receive version A is 0.0124

So, in percent the probability is 0.0124 × 100 = 1.24%

To the nearest tenth, the required probability is 1.2%.

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