The third term of a GP is 3. What is the product of the first five terms?
243
Calculating the Product of Terms in a Geometric Progression (GP)
This problem involves finding the product of the first five terms of a Geometric Progression (GP), given the value of its third term. Let's break down how to solve this.
Understanding Geometric Progressions
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 denoted by $a$ and the common ratio by $r$, 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: \(a_5 = ar^4\)
In general, the $n$-th term of a GP is given by \(a_n = ar^{n-1}\).
Given Information
We are given that the third term of the GP is 3. Using the formula for the $n$-th term, the third term ($n=3$) is:
\(a_3 = ar^{3-1} = ar^2\)
So, we have the equation:
\(ar^2 = 3\)
Finding the Product of the First Five Terms
We need to find the product of the first five terms of the GP. Let \(P_5\) be this product.
\(P_5 = a_1 \times a_2 \times a_3 \times a_4 \times a_5\)
Substituting the terms in terms of $a$ and $r$:
\(P_5 = a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)\)
Now, let's group the terms with $a$ and the terms with $r$:
\(P_5 = (a \times a \times a \times a \times a) \times (r^0 \times r^1 \times r^2 \times r^3 \times r^4)\)
\(P_5 = a^5 \times r^{(0+1+2+3+4)}\)
The sum of the exponents of $r$ is $0+1+2+3+4 = 10$.
\(P_5 = a^5 r^{10}\)
We can rewrite this expression by grouping $a$ and \(r^2\) together:
\(P_5 = (a r^2)^5\)
Using the Given Information to Calculate the Product
We know that \(ar^2 = 3\). We can substitute this value into the expression for \(P_5\):
\(P_5 = (3)^5\)
Now, we calculate the value of \(3^5\):
\(3^5 = 3 \times 3 \times 3 \times 3 \times 3\)
\(3^5 = 9 \times 9 \times 3\)
\(3^5 = 81 \times 3\)
\(3^5 = 243\)
Thus, the product of the first five terms of the GP is 243.
Conclusion
Given that the third term of a Geometric Progression is 3, the product of its first five terms is 243. The structure of the product of the first \(n\) terms of a GP often simplifies nicely when an intermediate term is known.
| Term Number (n) | Term (\(a_n\)) |
|---|---|
| 1 | \(a\) |
| 2 | \(ar\) |
| 3 | \(ar^2 = 3\) |
| 4 | \(ar^3\) |
| 5 | \(ar^4\) |
Revision Table: Geometric Progression Concepts
| Concept | Description | Formula |
|---|---|---|
| First term | The initial term of the sequence. | \(a\) |
| Common ratio | The constant factor between consecutive terms. | \(r = \frac{a_{n+1}}{a_n}\) |
| \(n\)-th term | The term at position \(n\). | \(a_n = ar^{n-1}\) |
| Product of first \(n\) terms | The product \(a_1 \times a_2 \times \dots \times a_n\). | \(P_n = a^n r^{n(n-1)/2}\) |
Additional Information: Product of GP Terms
For a GP with first term \(a\) and common ratio \(r\), the product of the first \(n\) terms is \(P_n = a \cdot ar \cdot ar^2 \cdot \dots \cdot ar^{n-1}\). This simplifies to \(P_n = a^n \cdot r^{0+1+2+\dots+(n-1)}\).
The sum of the exponents of \(r\) is an arithmetic series sum: \(0+1+2+\dots+(n-1) = \frac{(n-1)n}{2}\).
So, the general formula for the product of the first \(n\) terms is \(P_n = a^n r^{n(n-1)/2}\).
In our case, \(n=5\). The formula gives \(P_5 = a^5 r^{5(5-1)/2} = a^5 r^{5(4)/2} = a^5 r^{10}\). As shown in the solution, this can be conveniently written as \((ar^2)^5\).
This demonstrates a useful property: the product of the first \(n\) terms of a GP can often be related to the middle term or geometric mean of the terms, especially when \(n\) is odd. For \(n=5\), the third term \(ar^2\) is the geometric mean of the first and fifth terms (\(\sqrt{a \cdot ar^4} = \sqrt{a^2 r^4} = ar^2\)) and also the geometric mean of the second and fourth terms (\(\sqrt{ar \cdot ar^3} = \sqrt{a^2 r^4} = ar^2\)). The product \(P_5\) is simply the third term raised to the power of 5, i.e., \((ar^2)^5\).
Similar Questions
The fifth term of an AP of n terms, whose sum is n 2– 2n, is
A person is to count 4500 notes. Let a ndenote the number of notes he counts in the nth minute. If a 1= a 2= a 3= … = a 10 = 150, and a 10 , a 11 , a 12 , … are in AP with the common difference -2, then the time taken by him to count all the notes is
If \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) , where S ndenotes the sum of the first n terms of an AP, then the common difference is
If the ratio of AM to GM of two positive numbers a and b is 5 : 3 then a : b is equal to
Let x, y, z be positive real numbers such that x, y, z are in GP and tan -1 x, tan -1 y and tan -1 z are in AP. Then which one of the following is correct?
If x 1and x 2are positive quantities, then the condition for the difference between the arithmetic mean and the geometric mean to be greater than 1 is
If y = x + x 2+ x 3+ … up to infinite terms where x < 1, then which one of the following is correct?
What is the sum of all two-digit numbers which when divided by 3 leave 2 as the remainder?
If x, 3/2, z are in AP; x, 3, z are in GP; then which one of the following will be in HP?
Consider the following statements :
1. 2 + 4 + 6 + ........ + 2n = n 2 + n
2. The expression n 2 + n + 41 always gives a prime number for every natural number n
Which of the above statements is/are correct ?
Important Questions from Sequences and Series
The fifth term of an AP of n terms, whose sum is n 2– 2n, is
A person is to count 4500 notes. Let a ndenote the number of notes he counts in the nth minute. If a 1= a 2= a 3= … = a 10 = 150, and a 10 , a 11 , a 12 , … are in AP with the common difference -2, then the time taken by him to count all the notes is
- \(\mathop {\lim }\limits_{n \to \infty } {\left( {1 - \frac{1}{{2n}}} \right)^{n + 1}}\) is equal to
If \({S_n} = nP + \frac{{n\left( {n - 1} \right)Q}}{2}\) , where S ndenotes the sum of the first n terms of an AP, then the common difference is
What is the sum of the first 12 terms of an arithmetic progression if the 3rd term is -13 and the 6th term is -4?
