Question

Insert three geometric means between 2 and 81/8

Answer 1

Answer:

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 2

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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