What is the product of first 2n+1 terms of a geometric progression ?

Options

A. The (n+1)th power of the nth term of the GP
B. The (2n+1)th power of the nth term of the GP
C. The (2n+1) the powerr of the (n+1)th term of the GP
D. The nth power of the (n+1)th terms of the GP

Answer

To find the product of the first \(2n + 1\) terms of a geometric progression (GP), we can follow these steps:

### Step 1: Understand the terms of a GP
A geometric progression is defined by its first term \(a\) and a common ratio \(r\). The terms of the GP can be expressed as:
- First term: \(a\)
- Second term: \(ar\)
- Third term: \(ar^2\)
- ...
- \(k\)-th term: \(ar^{k-1}\)

For the first \(2n + 1\) terms, the terms will be:
- \(a, ar, ar^2, ar^3, \ldots, ar^{2n}\)

### Step 2: Write the product of the first \(2n + 1\) terms
The product \(P\) of the first \(2n + 1\) terms is given by:
\[ P = a \cdot ar \cdot ar^2 \cdots ar^{2n} \]

### Step 3: Factor out common terms
We can factor out \(a\) from each term:
\[ P = a^{2n + 1} \cdot (r^0 \cdot r^1 \cdot r^2 \cdots r^{2n}) \]

### Step 4: Simplify the product of \(r\) terms
The product of the \(r\) terms can be simplified using the formula for the sum of an arithmetic series. The exponent of \(r\) is the sum of the first \(2n\) natural numbers:
\[ 0 + 1 + 2 + \ldots + 2n = \frac{(2n)(2n + 1)}{2} = n(2n + 1) \]
Thus, we can rewrite the product as:
\[ P = a^{2n + 1} \cdot r^{n(2n + 1)} \]

### Step 5: Rewrite in terms of the \(n\)-th and \((n + 1)\)-th term
The \(n\)-th term of the GP is given by:
\[ T_n = ar^{n - 1} \]
The \((n + 1)\)-th term is:
\[ T_{n+1} = ar^n \]

### Step 6: Express the product in terms of the \(n\)-th term
We can express \(P\) in terms of the \(n\)-th term:
\[ P = (ar^n)^{2n + 1} = T_{n+1}^{2n + 1} \]

### Conclusion
Thus, the product of the first \(2n + 1\) terms of a geometric progression is:
\[ P = T_{n+1}^{2n + 1} \]

### Final Answer
The product of the first \(2n + 1\) terms of a geometric progression is \(T_{n+1}^{2n + 1}\).

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