Sure! Let's delve into the given Geometric Progression (G.P.) with the terms: [tex]\(5, 25, 125\)[/tex].
### 1. Finding the Sum to Infinity
We start with the first term [tex]\(a = 5\)[/tex] and the common ratio [tex]\( r \)[/tex].
The common ratio [tex]\( r \)[/tex] is calculated as:
[tex]\[ r = \frac{\text{second term}}{\text{first term}} = \frac{25}{5} = 5 \][/tex]
In general, the sum to infinity for a G.P. is given by the formula:
[tex]\[ S_{\infty} = \frac{a}{1 - r} \][/tex]
However, this formula is valid only if the absolute value of the common ratio [tex]\(|r|\)[/tex] is less than 1. In this case, [tex]\( |r| = 5 \)[/tex], which is not less than 1. Therefore, the sum to infinity for this G.P. is:
[tex]\[ S_{\infty} = \text{undefined} \][/tex]
### 2. Finding the 12th Term
The nth term of a G.P. is given by:
[tex]\[ a_n = a \cdot r^{(n-1)} \][/tex]
For the 12th term ([tex]\(n = 12\)[/tex]):
[tex]\[ a_{12} = 5 \cdot 5^{(12-1)} = 5 \cdot 5^{11} = 5 \cdot 244140625 = 244140625 \][/tex]
Thus, the 12th term is:
[tex]\[ a_{12} = 244140625 \][/tex]
### 3. Finding the Sum of the First 16 Terms
The sum of the first [tex]\(n\)[/tex] terms of a G.P. is given by:
[tex]\[ S_n = a \cdot \frac{1 - r^n}{1 - r} \quad \text{if} \quad r \neq 1 \][/tex]
For the first 16 terms ([tex]\(n = 16\)[/tex]):
[tex]\[ S_{16} = 5 \cdot \frac{1 - 5^{16}}{1 - 5} \][/tex]
Calculating the power and the sum:
[tex]\[ S_{16} = 5 \cdot \frac{1 - 152587890625}{-4} = 5 \cdot \frac{-152587890624}{-4} = 5 \cdot 38146972656 = 190734863280 \][/tex]
Therefore, the sum of the first 16 terms is:
[tex]\[ S_{16} = 190734863280 \][/tex]
### Summary of Results
- Sum to infinity: Undefined
- 12th term: [tex]\( 244140625 \)[/tex]
- Sum of the first 16 terms: [tex]\( 190734863280 \)[/tex]
If you have any further questions or need additional explanations, feel free to ask!
