Mathematics, 19.05.2020 16:04 lovelyashjoy

Insert three geometric means between 2 and 81/8

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Answer from: thiago22

In a geometric sequence where three terms lie between 2 and 81/8, the three geometric terms are:

$\mathbf{T_2 =3 }$

$\mathbf{T_3 =\frac{9}{2} }$

$\mathbf{T_4 =\frac{27}{4} }$

Recall:

nth term of a geometric sequence is given as: $\mathbf{T_n = ar^{n - 1}}$ a = the first term; r = the common ratio; n = the number of terms

Given a geometric sequence, 2 . . . 81/8, with three other terms in the middle, first, find the value of r.

Thus:

First Term:

a = 2

Fifth Term:

$T_5 = ar^{n - 1}$

a = 2

n = 5

r = ?

T5 = 81/8

Plug in the value of a, n, and T5

$\frac{81}{8} = 2r^{5 - 1}\\\\\frac{81}{8} = 2r^4\\\\$

Multiply both sides by 8

$\frac{81}{8} \times 8 = 2r^4 \times 8\\\\81 = 16r^4\\\\$

Divide both sides by 16

$\frac{81}{16} = \frac{16r^4}{16} \\\\\frac{81}{16} = r^4\\\\$

Take the fourth root of both sides

$\sqrt[4]{\frac{81}{16}} = r\\\\\frac{3}{2} = r\\\\\mathbf{r = \frac{3}{2}}$

Find the three geometric means T_2, T_3, $ and $ T_4 between 2 and 81/8.

$\mathbf{T_n = ar^{n - 1}}$

a = 2

r = 3/2

Thus:

$T_2 = 2 \times (\frac{3}{2}) ^{2 - 1}\\\\T_2 = 2 \times (\frac{3}{2}) ^{1}\\\\\mathbf{T_2 = 3}$

$T_3 = 2 \times \frac{3}{2} ^{3 - 1}\\\\T_3 = 2 \times (\frac{3}{2}) ^{2}\\\\T_3 = 2 \times \frac{9}{4}\\\\\mathbf{T_3 =\frac{9}{2} }$

$T_4 = 2 \times \frac{3}{2} ^{4 - 1}\\\\T_4 = 2 \times (\frac{3}{2}) ^{3}\\\\T_4 = 2 \times \frac{27}{8}\\\\\mathbf{T_4 =\frac{27}{4} }$

Therefore, in a geometric sequence where three terms lie between 2 and 81/8, the three geometric terms are:

$\mathbf{T_2 =3 }$

$\mathbf{T_3 =\frac{9}{2} }$

$\mathbf{T_4 =\frac{27}{4} }$

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Answer from: lalllda

The three geometric means are 3, 9/2 and 27/4

Step-by-step explanation:

The nth term of a geometric sequence is expressed as Tn = $ar^{n-1}$ where;

a is the first term

r is the common ratio

n is the number of terms

Since we are to insert three geometric means between 2 and 81/8, the total number of terms in the sequence will be 5 terms as shown;

2, a, b, c, 81/8

a, b, and c are the 3 geometric mean to be inserted

T1 = $ar^{1-1}$ = 2

T1 = a = 2....(1)

T5= $ar^{5-1}$

T5 = $ar^{4}$ = 81/8... (2)

Dividing equation 1 by 2 we have;

$\frac{ar^{4} }{a}= \frac{\frac{81}{8} }{2}$

$r^{4} = \frac{81}{16}\\\\r = \sqrt[4]{\frac{81}{16} } \\r = 3/2$

Given a =2 and r = 3/2;

$T2=ar\\T2 = 2*3/2\\T2 = 3\\\\T3 = ar^{2} \\T3 = 2*\frac{3}{2} ^{2} \\T3 = 2*9/4\\T3 = 9/2\\\\T4 = ar^{3}\\T4 = 2*\frac{3}{2} ^{3} \\T4 = 2*27/8\\T4 = 27/4\\$

Therefore the three geometric means are 3, 9/2 and 27/4

Answer from: Quest

we'll this would mean that the dunk tank and the bounce house are the same for 20 hours because the 2 represents that the two activities, and the 20 represents the amount of hours they both are. so with this said, it means that the answer is c. the costs for a dunk tank and bounce house are the same for 20 hours.

c.the costs for a dunk tank and bounce house are the same for 20 hours.

hope this .

Answer from: Quest

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)

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