Certainly! Let's tackle this problem step by step.
### Given:
Three consecutive terms of a geometric progression (GP) are:
1. [tex]\(3^{\alpha x + 1}\)[/tex]
2. [tex]\(9^\alpha\)[/tex]
3. [tex]\(3^{\alpha (x + 1)}\)[/tex]
### a) Finding the value of [tex]\(x\)[/tex]:
In a geometric progression, the ratio between consecutive terms remains constant. That is, the ratio of the second term to the first term should be equal to the ratio of the third term to the second term:
[tex]\[
\frac{9^\alpha}{3^{\alpha x + 1}} = \frac{3^{\alpha (x + 1)}}{9^\alpha}
\][/tex]
First, let's express [tex]\(9^\alpha\)[/tex] in terms of base 3:
[tex]\[
9^\alpha = (3^2)^\alpha = 3^{2\alpha}
\][/tex]
Now, we can write the ratios:
[tex]\[
\frac{3^{2\alpha}}{3^{\alpha x + 1}} = \frac{3^{\alpha (x + 1)}}{3^{2\alpha}}
\][/tex]
Simplify the ratios by subtracting the exponents:
[tex]\[
\frac{3^{2\alpha}}{3^{\alpha x + 1}} = 3^{2\alpha - (\alpha x + 1)} = 3^{2\alpha - \alpha x - 1}
\][/tex]
[tex]\[
\frac{3^{\alpha (x + 1)}}{3^{2\alpha}} = 3^{\alpha (x + 1) - 2\alpha} = 3^{\alpha x + \alpha - 2\alpha} = 3^{\alpha x - \alpha}
\][/tex]
Now, because the common ratio must be the same:
[tex]\[
3^{2\alpha - \alpha x - 1} = 3^{\alpha x - \alpha}
\][/tex]
Since the bases (3) are the same, we can set the exponents equal to each other:
[tex]\[
2\alpha - \alpha x - 1 = \alpha x - \alpha
\][/tex]
Combine like terms:
[tex]\[
2\alpha + \alpha = \alpha x + \alpha x - 1
\][/tex]
[tex]\[
3\alpha - 1 = 2\alpha x
\][/tex]
Simplifying:
[tex]\[
3\alpha - 1 = \alpha x
\][/tex]
[tex]\[
x = \frac{3\alpha - 1}{\alpha}
\][/tex]
### b) Finding the common ratio:
Substitute the value of [tex]\(x\)[/tex] back into the ratio between the first and second terms:
[tex]\[
\text{Common Ratio (CR)} = \frac{9^\alpha}{3^{\alpha x + 1}} = \frac{3^{2\alpha}}{3^{\alpha \left( \frac{3\alpha - 1}{\alpha} \right) + 1}}
\][/tex]
Simplify the exponent:
[tex]\[
\alpha \left( \frac{3\alpha - 1}{\alpha} \right) + 1 = 3\alpha - 1 + 1 = 3\alpha
\][/tex]
So:
[tex]\[
\frac{3^{2\alpha}}{3^{3\alpha}} = 3^{2\alpha - 3\alpha} = 3^{-\alpha}
\][/tex]
Thus, the common ratio [tex]\(r\)[/tex] is:
[tex]\[
r = 3^{-\alpha}
\][/tex]
### c) Calculating the sum:
The three terms are:
1. [tex]\(3^{\alpha x + 1}\)[/tex]
2. [tex]\(9^\alpha = 3^{2\alpha}\)[/tex]
3. [tex]\(3^{\alpha (x+1)}\)[/tex]
Convert the terms by substituting [tex]\(x = \frac{3\alpha - 1}{\alpha}\)[/tex]:
1. [tex]\(3^{\alpha \left( \frac{3\alpha - 1}{\alpha} \right) + 1} = 3^{3\alpha - 1 + 1} = 3^{3\alpha}\)[/tex]
2. [tex]\(3^{2\alpha}\)[/tex]
3. [tex]\(3^{\alpha ( \frac{3\alpha - 1}{\alpha} + 1)} = 3^{3\alpha}\)[/tex]
Therefore, the sum of the terms:
[tex]\[
\text{Sum} = 3^{3\alpha} + 3^{2\alpha} + 3^{3\alpha} = 3^{3\alpha} + 3^{3\alpha} + 3^{2\alpha}
\][/tex]
[tex]\[
\text{Sum} = 2 \cdot 3^{3\alpha} + 3^{2\alpha}
\][/tex]
Thus, the stepwise solution provided gives us the value of [tex]\(x\)[/tex], the common ratio, and the sum of the geometric progression.
