Understanding Geometric Progression (GP) and the Problem

A Geometric Progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. If the terms are \(t_1, t_2, t_3, \ldots\), the general form of the \(n\)-th term is given by \(t_n = ar^{n-1}\), where \(a = t_1\) is the first term and \(r\) is the common ratio.

The problem asks for the value of the expression \({\left(t_1 t_3 \ldots t_{21}\right)}^{\frac{1}{11}}\) where \(t_1, t_2, t_3, \ldots\) are in GP.

Identifying the Terms in the Product

The terms in the product are the odd-indexed terms from \(t_1\) to \(t_{21}\). These terms are \(t_1, t_3, t_5, \ldots, t_{21}\).

Let's find the number of terms in this sequence of indices: \(1, 3, 5, \ldots, 21\). This is an arithmetic progression with first term 1, common difference 2, and last term 21. Let \(k\) be the number of terms.

The formula for the \(k\)-th term of an AP is \(a_k = a_1 + (k-1)d\).

So, \(21 = 1 + (k-1)2\)

\(20 = (k-1)2\)

\(10 = k-1\)

\(k = 11\)

There are 11 terms in the product.

Expressing the Terms in terms of 'a' and 'r'

Using the general formula \(t_n = ar^{n-1}\), we can write each term:

• \(t_1 = ar^{1-1} = ar^0 = a\)
• \(t_3 = ar^{3-1} = ar^2\)
• \(t_5 = ar^{5-1} = ar^4\)
• ...
• \(t_{21} = ar^{21-1} = ar^{20}\)

Calculating the Product

Let \(P = t_1 t_3 \ldots t_{21}\). Substituting the expressions in terms of 'a' and 'r':

\(P = (a) \times (ar^2) \times (ar^4) \times \ldots \times (ar^{20})\)

There are 11 terms, each containing 'a'. So, the power of 'a' in the product is 11.

The power of 'r' in the product is the sum of the exponents of 'r' from each term: \(0 + 2 + 4 + \ldots + 20\).

This is the sum of an arithmetic progression with first term 0, common difference 2, and 11 terms. The sum (S) is given by \(S = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term})\).

\(S = \frac{11}{2} \times (0 + 20)\)

\(S = \frac{11}{2} \times 20\)

\(S = 11 \times 10 = 110\)

So, the product \(P = a^{11} r^{110}\).

Evaluating the Expression

The expression we need to evaluate is \(P^{\frac{1}{11}}\).

\(P^{\frac{1}{11}} = {\left(a^{11} r^{110}\right)}^{\frac{1}{11}}\)

Using the property of exponents \((xy)^m = x^m y^m\) and \((x^p)^q = x^{pq}\):

\(P^{\frac{1}{11}} = (a^{11})^{\frac{1}{11}} \times (r^{110})^{\frac{1}{11}}\)

\(P^{\frac{1}{11}} = a^{11 \times \frac{1}{11}} \times r^{110 \times \frac{1}{11}}\)

\(P^{\frac{1}{11}} = a^1 \times r^{10}\)

\(P^{\frac{1}{11}} = ar^{10}\)

Relating the Result back to the GP Terms

Recall the general term of the GP is \(t_n = ar^{n-1}\). Comparing this with our result \(ar^{10}\), we can see that \(ar^{10} = ar^{11-1} = t_{11}\).

Conclusion

The value of \({\left(t_1 t_3 \ldots t_{21}\right)}^{\frac{1}{11}}\) is equal to \(t_{11}\).

Therefore, the correct option is \(t_{11}\).