Answer :
To solve this problem, we need to find the common ratio [tex]\( r \)[/tex], the first term [tex]\( a \)[/tex], and the sum of the first 6 terms of a geometric progression (GP). Given the information:
1. The 3rd term: [tex]\( a \cdot r^2 = \text{los} \)[/tex] (which we'll handle assuming there's a typo, leaving it as a variable [tex]\( x \)[/tex])
2. The 6th term: [tex]\( a \cdot r^5 = -32 \)[/tex]
### Step-by-Step Solution
#### a) Find the common ratio ([tex]\( r \)[/tex])
We have:
[tex]\[ a \cdot r^5 = -32 \][/tex]
From the 3rd term (where [tex]\( a \cdot r^2 = x \)[/tex]), we can't directly use the value due to the apparent typo ("los"). So we'll continue first solving [tex]\( r \)[/tex] assuming [tex]\( x \)[/tex] is a numeric placeholder.
Rewriting [tex]\( a \)[/tex] from the 3rd term equation:
[tex]\[ a = \frac{x}{r^2} \][/tex]
Put this value of [tex]\( a \)[/tex] into the 6th term equation:
[tex]\[ \left( \frac{x}{r^2} \right) \cdot r^5 = -32 \][/tex]
[tex]\[ x \cdot r^3 = -32 \][/tex]
[tex]\[ r^3 = \frac{-32}{x} \][/tex]
Taking the cube root:
[tex]\[ r = \sqrt[3]{\frac{-32}{x}} \][/tex]
#### b) Find the first term ([tex]\( a \)[/tex])
Now that we have [tex]\( r \)[/tex], we can find [tex]\( a \)[/tex] using the 3rd term equation:
[tex]\[ a \cdot r^2 = x \][/tex]
Thus:
[tex]\[ a = \frac{x}{r^2} \][/tex]
Substitute [tex]\( r = \sqrt[3]{\frac{-32}{x}} \)[/tex] into this:
[tex]\[ r^2 = \left(\sqrt[3]{\frac{-32}{x}}\right)^2 = \left(\frac{-32}{x}\right)^{2/3} \][/tex]
Then:
[tex]\[ a = \frac{x}{\left(\frac{-32}{x}\right)^{2/3}} = x \cdot \left(\frac{x}{-32}\right)^{2/3} = x \cdot \frac{x^{2/3}}{(-32)^{2/3}} = x^{1 + 2/3} \cdot \frac{1}{(-32)^{2/3}} = \frac{x^{5/3}}{(-32)^{2/3}} \][/tex]
#### c) Find the sum of the first 6 terms
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a GP is given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \][/tex]
For [tex]\( n = 6 \)[/tex], using our values for [tex]\( a \)[/tex] and [tex]\( r \)[/tex]:
[tex]\[ S_6 = \left(\frac{x^{5/3}}{(-32)^{2/3}}\right) \cdot \frac{1 - \left(\sqrt[3]{\frac{-32}{x}}\right)^6}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Simplify:
[tex]\[ S_6 = \left(\frac{x^{5/3}}{16 \cdot (-1)^{2/3}}\right) \cdot \frac{1 - \frac{(-32)^2}{x^2}}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Since [tex]\((-1)^{2/3} = 1\)[/tex] and [tex]\((-32)^{2/3} = 16\)[/tex]:
[tex]\[ S_6 = \frac{x^{5/3}}{16} \cdot \frac{1 - \frac{1024}{x^2}}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Combine terms inside the fraction:
[tex]\[ S_6 = \frac{x^{5/3}}{16} \cdot \frac{\frac{x^2 - 1024}{x^2}}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Therefore, the common ratio [tex]\( r \)[/tex], the first term [tex]\( a \)[/tex], and the sum of the first 6 terms for the given GP involving [tex]\( x \)[/tex] as placeholder for the corrected 3rd term value are derived. Necessary numeric calculations can process accurately upon the establishment of [tex]\( x \)[/tex].
1. The 3rd term: [tex]\( a \cdot r^2 = \text{los} \)[/tex] (which we'll handle assuming there's a typo, leaving it as a variable [tex]\( x \)[/tex])
2. The 6th term: [tex]\( a \cdot r^5 = -32 \)[/tex]
### Step-by-Step Solution
#### a) Find the common ratio ([tex]\( r \)[/tex])
We have:
[tex]\[ a \cdot r^5 = -32 \][/tex]
From the 3rd term (where [tex]\( a \cdot r^2 = x \)[/tex]), we can't directly use the value due to the apparent typo ("los"). So we'll continue first solving [tex]\( r \)[/tex] assuming [tex]\( x \)[/tex] is a numeric placeholder.
Rewriting [tex]\( a \)[/tex] from the 3rd term equation:
[tex]\[ a = \frac{x}{r^2} \][/tex]
Put this value of [tex]\( a \)[/tex] into the 6th term equation:
[tex]\[ \left( \frac{x}{r^2} \right) \cdot r^5 = -32 \][/tex]
[tex]\[ x \cdot r^3 = -32 \][/tex]
[tex]\[ r^3 = \frac{-32}{x} \][/tex]
Taking the cube root:
[tex]\[ r = \sqrt[3]{\frac{-32}{x}} \][/tex]
#### b) Find the first term ([tex]\( a \)[/tex])
Now that we have [tex]\( r \)[/tex], we can find [tex]\( a \)[/tex] using the 3rd term equation:
[tex]\[ a \cdot r^2 = x \][/tex]
Thus:
[tex]\[ a = \frac{x}{r^2} \][/tex]
Substitute [tex]\( r = \sqrt[3]{\frac{-32}{x}} \)[/tex] into this:
[tex]\[ r^2 = \left(\sqrt[3]{\frac{-32}{x}}\right)^2 = \left(\frac{-32}{x}\right)^{2/3} \][/tex]
Then:
[tex]\[ a = \frac{x}{\left(\frac{-32}{x}\right)^{2/3}} = x \cdot \left(\frac{x}{-32}\right)^{2/3} = x \cdot \frac{x^{2/3}}{(-32)^{2/3}} = x^{1 + 2/3} \cdot \frac{1}{(-32)^{2/3}} = \frac{x^{5/3}}{(-32)^{2/3}} \][/tex]
#### c) Find the sum of the first 6 terms
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of a GP is given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \][/tex]
For [tex]\( n = 6 \)[/tex], using our values for [tex]\( a \)[/tex] and [tex]\( r \)[/tex]:
[tex]\[ S_6 = \left(\frac{x^{5/3}}{(-32)^{2/3}}\right) \cdot \frac{1 - \left(\sqrt[3]{\frac{-32}{x}}\right)^6}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Simplify:
[tex]\[ S_6 = \left(\frac{x^{5/3}}{16 \cdot (-1)^{2/3}}\right) \cdot \frac{1 - \frac{(-32)^2}{x^2}}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Since [tex]\((-1)^{2/3} = 1\)[/tex] and [tex]\((-32)^{2/3} = 16\)[/tex]:
[tex]\[ S_6 = \frac{x^{5/3}}{16} \cdot \frac{1 - \frac{1024}{x^2}}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Combine terms inside the fraction:
[tex]\[ S_6 = \frac{x^{5/3}}{16} \cdot \frac{\frac{x^2 - 1024}{x^2}}{1 - \sqrt[3]{\frac{-32}{x}}} \][/tex]
Therefore, the common ratio [tex]\( r \)[/tex], the first term [tex]\( a \)[/tex], and the sum of the first 6 terms for the given GP involving [tex]\( x \)[/tex] as placeholder for the corrected 3rd term value are derived. Necessary numeric calculations can process accurately upon the establishment of [tex]\( x \)[/tex].
