π Example 1.

Insert two numbers between 3 and 81 so that the resulting sequence is geometric sequence.

β Solution:

Let *G*_{1} and *G*_{2} be two numbers between 3 and 81 such that the series, 3, *G*_{1}, *G*_{2}, 81, forms a geometric sequence.

Let a be the first term and *r* be the common ratio of the geometric sequence.

*r*

^{3}

*r*

^{3}=27

*r*=3

*G*

_{1}=

*a*

_{2}=

*ar*=3β 3=9,

*G*

_{2}=

*a*

_{3}=

*ar*

^{2}=3β 3

^{2}=27.

Thus, the required two numbers are 9 and 27.

π Ex2. **Find the indicated geometric means between two terms**.

π Ex2a. 0.20, _, _, _, 125

β Solution: Given *a*_{1}=0.20 and *a*_{5}=125.

*a*=

_{n}*a*

_{1}β

*r*

^{(n-1)}

125=0.20β

*r*

^{(5-1)}

*r*

^{4}=625

*r*=Β±5

The geometric means are 1, 5, 25 or -1, 5, -25.

Answer: 1, 5, 25 or -1, 5, -25

π Ex2b. 4 and 256; 2 means

β Solution:

The sequence will resemble 4, ? , ? , 256.

Note that *a*_{1}=4, *n*=4, and *a*_{4}=256. Find the common ratio using *n*th term for a geometric sequence formula.

*a*=

_{n}*a*

_{1}β

*r*

^{(n-1)}

256=4

*r*

^{3}

64=

*r*

^{3}

4=

*r*

The common ratio is 4. Use

*r*to find the geometric means.

16(4)=64

Therefore, a sequence with two geometric means between 4 and 256 is 4, 16, 64, 256.

π Ex2c. 256 and 81; 3 means

β Solution:

The sequence will resemble 256, ? , ? , ? , 81. Note that *a*_{1}=256, *n*=5, and *a*_{5}=81.

Find the common ratio using the *n*th term for a geometric sequence formula.

The common ratio is Β±ΒΎ. Use

*r*to find the geometric means.

*r*=-ΒΎ

256β (-ΒΎ)=-192

-192β (-ΒΎ)=144

144β (-ΒΎ)=108

*r*=ΒΎ

256β (ΒΎ)=192

192β (ΒΎ)=144

144β (ΒΎ)=108

Therefore, a sequence with three geometric means between 256 and 81 is 256, -192, 144, -108, 81 or 256, 192, 144, 108, 81.

π Ex2d. 4/7 and 7; 1 mean

β Solution:

The sequence will resemble 4/7, ? , 7. Note that *a*_{1}=4/7, *n*=3, and *a*_{3}=7. Find the common ratio using the *n*th term for a geometric sequence formula.

The common ratio is Β±7/2 . Use

*r*to find the geometric means.

Therefore, a sequence with one geometric mean between 4/7 and 7 is 4/7, -2 , 7 or 4/7, 2 , 7

π Ex2e. -2 and 54; 2 means

β Solution:

The sequence will resemble -2, ? , ? , 54.

Note that *a*_{1}=-2, *n*=4, and *a*_{4}=54. Find the common ratio using *n*th term for a geometric sequence formula.

*a*=

_{n}*a*

_{1}β

*r*

^{(n-1)}

54=-2

*r*

^{3}

-27=

*r*

^{3}

-3=

*r*

The common ratio is -3. Use

*r*to find the geometric means.

6(-3)=-18

Therefore, a sequence with two geometric means between -2 and 54 is -2, 6, -18, 54.

π Ex2f. 1 and 27; 2 means

β Solution: The sequence will resemble 1, ? , ? , 27.

Note that *a*_{1}=1, *n*=4, and *a*_{4}=27. Find the common ratio using *n*th term for a geometric sequence formula.

*a*=

_{n}*a*

_{1}β

*r*

^{(n-1)}

27=1

*r*

^{3}

3

^{3}=

*r*

^{3}

3=

*r*

The common ratio is 3. Use

*r*to find the geometric means.

3(3)=9

Therefore, a sequence with two geometric means between 1 and 27 is 1, 3, 9, 27.

π Ex2g. 48 and -750; 2 means

β Solution:

The sequence will resemble 48, ? , ? , -750.

Note that *a*_{1}=48, *n*=4, and *a*_{4}=48. Find the common ratio using *n*th term for a geometric sequence formula.

The common ratio is -5/2. Use

*r*to find the geometric means.

Therefore, a sequence with two geometric means between 48 and -750 is 48, -120, 300, -750.

π Ex2h. *t*^{8} and *t*^{(-7)}; 4 means

β Solution:

The sequence will resemble *t*^{8}, ? , ? , ? , ? , *t*^{(-7)}.

Note that *a*_{1}=*t*^{8}, *n*=6, and *a*_{6}=*t*^{(-7)}. Find the common ratio using *n*th term for a geometric sequence formula.

The common ratio is

*t*

^{(-3)}. Use

*r*to find the geometric means.

*t*

^{8}β

*t*

^{(-3)}=

*t*

^{5}

*t*

^{5}β

*t*

^{(-3)}=

*t*

^{2}

*t*

^{2}β

*t*

^{(-3)}=

*t*

^{(-1)}

*t*

^{(-1)}β

*t*

^{(-3)}=

*t*

^{(-4)}

Therefore, a sequence with two geometric means between

*t*

^{8}and

*t*

^{(-7)}is

*t*

^{8},

*t*

^{5},

*t*

^{2},

*t*

^{(-1)},

*t*

^{(-4)},

*t*

^{(-7)}.