To solve this problem, we need to determine the number of geometric means inserted between 4 and 128 such that the ratio of the first mean to the last mean is 1:8. Here's a step-by-step solution:
1. Identify the first and last terms:
- First term ([tex]\( a \)[/tex]) = 4
- Last term ([tex]\( b \)[/tex]) = 128
2. Determine the number of terms in the sequence:
- Let [tex]\( n \)[/tex] be the number of geometric means we need to find. Therefore, the total number of terms in the sequence, including the first and last terms, is [tex]\( n + 2 \)[/tex].
3. Formula for the [tex]\( k \)[/tex]-th term in a geometric sequence:
- The general form for the [tex]\( k \)[/tex]-th term [tex]\( T_k \)[/tex] in a geometric sequence is [tex]\( T_k = ar^{k-1} \)[/tex], where [tex]\( a \)[/tex] is the first term and [tex]\( r \)[/tex] is the common ratio.
4. Find the common ratio [tex]\( r \)[/tex]:
- The last term [tex]\( b \)[/tex] is the [tex]\((n+2)\)[/tex]-th term of the sequence.
- Therefore, [tex]\( 128 = 4 \cdot r^{(n+2)-1} = 4 \cdot r^{n+1} \)[/tex].
- Solving for [tex]\( r \)[/tex]:
[tex]\[
128 = 4r^{n+1} \implies r^{n+1} = \frac{128}{4} = 32 \implies r = 32^{\frac{1}{n+1}}
\][/tex]
5. Use the ratio of the first and last mean:
- The first mean (M1) is the second term of the sequence (since it comes after the first term).
- Therefore, [tex]\( M1 = 4r \)[/tex].
- The last mean (Mn) is the [tex]\( (n+2-1) \)[/tex]-th term of the sequence.
- Therefore, [tex]\( Mn = 4r^n \)[/tex].
The ratio of the first mean (M1) to the last mean (Mn) is given by:
[tex]\[
\frac{M1}{Mn} = \frac{4r}{4r^n} = \frac{r}{r^n} = \frac{1}{r^{n-1}}
\][/tex]
And this ratio is equal to [tex]\( \frac{1}{8} \)[/tex]:
[tex]\[
\frac{1}{r^{n-1}} = \frac{1}{8} \implies r^{n-1} = 8
\][/tex]
6. Combine the relationships:
- We know from the previous steps that [tex]\( r = 32^{\frac{1}{n+1}} \)[/tex] and [tex]\( r^{n-1} = 8 \)[/tex].
Therefore:
[tex]\[
\left(32^{\frac{1}{n+1}}\right)^{n-1} = 8 \implies 32^{\frac{n-1}{n+1}} = 8
\][/tex]
7. Express 8 and 32 as powers of 2:
- Note that 32 can be written as [tex]\( 2^5 \)[/tex] and 8 can be written as [tex]\( 2^3 \)[/tex].
So:
[tex]\[
\left(2^5\right)^{\frac{n-1}{n+1}} = 2^3 \implies 2^{5 \cdot \frac{n-1}{n+1}} = 2^3
\][/tex]
- Equate the exponents of 2:
[tex]\[
5 \cdot \frac{n-1}{n+1} = 3 \implies \frac{5(n-1)}{n+1} = 3 \implies 5(n-1) = 3(n+1)
\][/tex]
8. Solve for [tex]\( n \)[/tex]:
- Distribute and combine like terms:
[tex]\[
5n - 5 = 3n + 3 \implies 5n - 3n = 3 + 5 \implies 2n = 8 \implies n = 4
\][/tex]
Therefore, the number of geometric means inserted between 4 and 128 is [tex]\( \boxed{4} \)[/tex].
