Solving the Geometric Progression (GP) Problem

The question asks us to find the product of the first five terms of a Geometric Progression (GP), given that its third term is 3.

Understanding Geometric Progression (GP)

A Geometric Progression is a sequence of non-zero 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 first term is \(a\) and the common ratio is \(r\), the terms of a GP are:

• First term: \(a_1 = a\)
• Second term: \(a_2 = ar\)
• Third term: \(a_3 = ar^2\)
• Fourth term: \(a_4 = ar^3\)
• Fifth term: \(a_5 = ar^4\)
• And so on, the \(n\)-th term is \(a_n = ar^{n-1}\).

Using the Given Information: The Third Term

We are given that the third term of the GP is 3. Using the formula for the \(n\)-th term, the third term (\(a_3\)) is \(ar^{3-1} = ar^2\).

So, we have the equation:

\(ar^2 = 3\)

Calculating the Product of the First Five Terms

We need to find the product of the first five terms, which we can denote as \(P_5\).

\(P_5 = a_1 \times a_2 \times a_3 \times a_4 \times a_5\)

Substitute the terms using the general form of a GP:

\(P_5 = a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)\)

Step-by-Step Calculation of the Product

Let's group the terms involving \(a\) and the terms involving \(r\):

\(P_5 = (a \times a \times a \times a \times a) \times (r \times r^2 \times r^3 \times r^4)\)

Using the rules of exponents (\(x^m \times x^n = x^{m+n}\)):

• The product of the 'a' terms is \(a^{1+1+1+1+1} = a^5\).
• The product of the 'r' terms is \(r^{1+2+3+4}\).

Let's calculate the sum of the exponents for \(r\):

\(1 + 2 + 3 + 4 = 10\)

So, the product of the 'r' terms is \(r^{10}\).

Therefore, the product of the first five terms is:

\(P_5 = a^5 r^{10}\)

We can rewrite this expression to make use of the given information \(ar^2 = 3\). Notice that \(a^5 r^{10}\) can be written as \((a r^2)^5\).

\(P_5 = (ar^2)^5\)

Now, substitute the value of \(ar^2\) which is 3:

\(P_5 = (3)^5\)

Finally, calculate the value of \(3^5\):

\(3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 3 \times 3 \times 3 = 27 \times 3 \times 3 = 81 \times 3 = 243\)

So, the product of the first five terms of the GP is 243.

Why the Third Term is Sufficient

In a GP, the terms are related by the common ratio. The product of terms symmetric about the middle term is constant. For the first five terms (\(a_1, a_2, a_3, a_4, a_5\)), the middle term is \(a_3\).

• \(a_1 \times a_5 = a \times ar^4 = ar^4\)
• \(a_2 \times a_4 = ar \times ar^3 = a^2 r^4 = (ar^2)^2 = 3^2 = 9\)
• \(a_3 = ar^2 = 3\)

Let's re-examine \(a_1 \times a_5\). Using \(ar^2 = 3\), we can write \(a = 3/r^2\). Then \(a_1 \times a_5 = (3/r^2) \times ar^4 = 3a r^2 = 3(3) = 9\). Alternatively, \(a_1 \times a_5 = a \times ar^4 = a^2 r^4 = (ar^2)^2 = 3^2 = 9\).

The product of the five terms is \(P_5 = a_1 \times a_2 \times a_3 \times a_4 \times a_5 = (a_1 \times a_5) \times (a_2 \times a_4) \times a_3\).

\(P_5 = (ar^4) \times (a^2r^4) \times (ar^2)\) -- this seems incorrect. Let's go back to \(P_5 = a^5 r^{10}\).

We showed \(P_5 = (ar^2)^5\). Since \(ar^2 = a_3\), the product is \(a_3^5\). This property holds for any odd number of terms in a GP: the product is the middle term raised to the power of the number of terms.

Product of first \(2n+1\) terms of a GP is \((a_{n+1})^{2n+1}\).

Here, \(2n+1 = 5\), so \(n=2\). The middle term is \(a_{2+1} = a_3\). The product is \((a_3)^5\).

Given \(a_3 = 3\), the product is \(3^5 = 243\). This confirms our earlier calculation and shows why only the third term was needed.

Revision Table: Key GP Concepts

Additional Information on Geometric Progression Properties

Geometric Progressions have several interesting properties:

• Relationship between terms: Any term is the geometric mean of its neighbors. For example, \(a_n = \sqrt{a_{n-1} \times a_{n+1}}\). For the five terms, \(a_3 = \sqrt{a_2 a_4}\) and \(a_3 = \sqrt{a_1 a_5}\).
• Product of equidistant terms: In a GP with an odd number of terms, the product of terms equidistant from the beginning and end is equal to the square of the middle term. For first five terms, \(a_1 \times a_5 = a \times ar^4 = a^2 r^4 = (ar^2)^2 = a_3^2\) and \(a_2 \times a_4 = ar \times ar^3 = a^2 r^4 = (ar^2)^2 = a_3^2\).
• Sum of a finite GP: The sum of the first \(n\) terms (\(S_n\)) is given by \(S_n = a \frac{1-r^n}{1-r}\) (if \(r \neq 1\)) or \(S_n = na\) (if \(r = 1\)).

The property regarding the product of equidistant terms being equal to the square of the middle term (for an odd number of terms) is particularly useful for problems like this one. If the number of terms is odd, say \(2n+1\), the product is the \((n+1)\)-th term raised to the power of \(2n+1\). Here, for 5 terms (\(2n+1=5\), so \(n=2\)), the middle term is the 3rd term (\(n+1=3\)), and the product is the 3rd term raised to the power of 5.