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3 years ago

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Mathematics

1 answer:

Zigmanuir [339]3 years ago

8 0

Answer:

Ai. Arithmetic sequence

Aii. Tn = 5 + 7n

Bi. Geometric

Bii. Tn = 8 × 2ⁿ¯¹

Step-by-step explanation:

To successfully answer the questions given above, note the following:

1. If the sequence is Arithmetic, then:

2nd – 1st = 3rd – 2nd = common difference (d)

2. If the sequence is geometric, then,

2nd / 1st = 3rd / 2nd = common ratio (r)

3. A sequence can not be arithmetic geometric at the same time.

4. The nth term of arithmetic sequence is:

Tn = a + (n – 1)d

5. The nth term of geometric sequence is:

Tn = arⁿ¯¹

A. Sequence => 12, 19, 26

i. Determination of the type of sequence.

We'll begin by calculating the common difference

1st term = 12

2nd term = 19

3rd term = 26

Common difference (d) = 2nd – 1st

d = 19 – 12 = 7

d = 3rd – 2nd

d = 26 – 19 = 7

Since a common difference exist in the sequence, the sequence is arithmetic sequence.

ii. Determination of the nth term.

Common difference (d) = 7

1st term (a) = 12

nth term (Tn) =?

Tn = a + (n – 1)d

Tn = 12 + (n – 1)7

Tn = 12 + 7n – 7

Tn = 5 + 7n

B. Sequence => 8, 16, 32

Bi. Determination of the type of sequence.

Let us begin by calculating the common ratio.

1st term = 8

2nd term = 16

3rd term = 32

Common ratio (r) = 2nd / 1st

r = 16 / 8

r = 2

r = 3rd / 2nd

r = 32 / 16

r = 2

Since a common ratio exist in the sequence, the sequence is geometric.

Bii. Determination of the nth term.

Common ratio(r) = 2

1st term (a) = 8

nth term =?

Tn = arⁿ¯¹

Tn = 8 × 2ⁿ¯¹

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Answer:

A. [1,10]

Step-by-step explanation:

Given

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Required

Interval with the fastest rate of decay

The rate of change over [a,b] is calculated as:

$Rate = \frac{N(b) - N(a)}{b - a}$

So, we have:

A. [1,10]

$Rate = \frac{N(10)-N(1)}{10 - 1}$

$Rate = \frac{N(10)-N(1)}{9}$

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$N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73$

So:

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B. [10,20]

$Rate = \frac{N(20)-N(10)}{20 - 10}$

$Rate = \frac{N(20)-N(10)}{10}$

Calculate N(20) and N(10)

$N(20) = 100e^{-0.023*20} = 100e^{-0.46} = 63.13$

$N(10) = 100e^{-0.023*10} = 100e^{-0.23} = 79.45$

So:

$Rate = \frac{63.13 - 79.45}{10} = \frac{-16.32}{10} = -1.632$

C. [15,25]

$Rate = \frac{N(25)-N(15)}{25 - 15}$

$Rate = \frac{N(25)-N(15)}{20}$

Calculate N(25) and N(15)

$N(25) = 100e^{-0.023*25} = 100e^{-0.575} = 56.27$

$N(15) = 100e^{-0.023*15} = 100e^{-0.345} = 70.82$

So:

$Rate = \frac{56.27 - 70.82}{10} = \frac{-14.55}{10} = -1.455$

D. [1,30]

$Rate = \frac{N(30)-N(1)}{30 - 1}$

$Rate = \frac{N(30)-N(1)}{29}$

Calculate N(30) and N(1)

$N(30) = 100e^{-0.023*30} = 100e^{-0.69} = 50.16$

$N(1) = 100e^{-0.023*1} = 100e^{-0.023} = 97.73$

So:

$Rate = \frac{50.16 - 97.73}{29} = \frac{-47.57}{29} = -1.64$

By comparing the values of the calculated rates,

-2.03 is the smallest, in other words; the fastest rate

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