Geometric Progression

Definition

A geometric progression (GP), or geometric sequence, 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. The recursive definition of a geometric progression can be described as follows:

Let \(a_1\) denote the first term of the geometric progression, and let \(r\) denote the common ratio, where \(r \neq 0\). Then, the \(n\)-th term of the geometric progression, denoted by \(a_n\), can be defined recursively by:

• \(a_1 = a\), where \(a\) is the first term of the sequence.
• For \(n > 1\), \(a_n = a_{n-1} \cdot r\), where \(a_{n-1}\) is the term immediately preceding \(a_n\), and \(r\) is the common ratio.

So, in a geometric progression (GP) with the first term \(a\) and common ratio \(r\), the sequence is: \(a\), \(ar\), \(ar^2\), \(ar^3\), ..., where each term is obtained by multiplying the previous term by \(r\).

For example: For a geometric progression (GP) with the first term \(a = 2\) and a common ratio \(r = \frac{1}{3}\), the sequence is: \(2\), \(2 \cdot \frac{1}{3}\), \(2 \cdot \left(\frac{1}{3}\right)^2\), \(2 \cdot \left(\frac{1}{3}\right)^3\), ..., resulting in the series: \(2\), \(\frac{2}{3}\), \(\frac{2}{9}\), \(\frac{2}{27}\), ....

The general term

To derive the \(n\)-th term of a geometric progression with the first term \(a\) and common ratio \(r\), we observe the relationship between successive terms:

\[ \frac{a_2}{a_1} = r, \quad \frac{a_3}{a_2} = r, \quad \frac{a_4}{a_3} = r, \ldots, \quad \frac{a_n}{a_{n-1}} = r \]

Multiplying these equations together, we get:

\[ \frac{a_2}{a_1} \cdot \frac{a_3}{a_2} \cdot \frac{a_4}{a_3} \cdot \ldots \cdot \frac{a_n}{a_{n-1}} = r \cdot r \cdot r \cdot \ldots \cdot r \]

Notice the telescoping effect, where all intermediate terms \(a_2, a_3, \ldots, a_{n-1}\) cancel out, leaving:

\[ \frac{a_n}{a} = r^{n-1} \]

Thus, we arrive at the general formula for the \(n\)-th term of the geometric progression:

\[ a_n = a \cdot r^{n-1} \]

This demonstrates that the \(n\)-th term is obtained by multiplying the first term \(a\) by the common ratio \(r\) raised to the power of \(n-1\), encapsulating the essence of geometric progression.

Increasing vs Decreasing GP

Consider the characteristics of the following geometric progressions (GPs), each defined by a distinct pair of initial term \(a\) and common ratio \(r\):

1. For \(a = 2\), \(r = 3\):
   The sequence begins as \(2, 6, 18, 54, \ldots\). Given \(a\) and \(r\) are positive, with \(r\) exceeding unity, the progression is monotonically increasing. Each term exceeds its predecessor, amplifying the sequence exponentially.

2. For \(a = 2\), \(r = \frac{1}{3}\):
   This progression, \(2, \frac{2}{3}, \frac{2}{9}, \frac{2}{27}, \ldots\), demonstrates a decrease in magnitude. The positive initial term combined with a common ratio between zero and one ensures a gradual reduction towards zero.

3. For \(a = -2\), \(r = 3\):
   Manifesting as \(-2, -6, -18, -54, \ldots\), this series, while consistently multiplying in magnitude due to a common ratio greater than one, starts with a negative term. The sequence, therefore, remains negative, diverging to negative infinity.

4. For \(a = -2\), \(r = \frac{1}{3}\):
   The series \(-2, -\frac{2}{3}, -\frac{2}{9}, -\frac{2}{27}, \ldots\) diminishes in absolute value. Originating from a negative term and reduced by a factor less than one, it converges towards zero from the negative side.

5. For \(a = 2\), \(r = -3\):
   With terms \(2, -6, 18, -54, \ldots\), the progression alternates in sign due to the negative common ratio. The sequence exhibits neither a pure increase nor decrease but expands in absolute magnitude, signifying an oscillatory behavior.

We can easily see that:

1. \(a > 0, r > 1\): The GP is monotonically increasing, as each term is larger than the previous one due to a positive starting term and a common ratio greater than one.
2. \(a > 0, 0 < r < 1\): The GP is monotonically decreasing, with each term getting closer to zero, starting from a positive first term and a common ratio between zero and one.
3. \(a < 0, r > 1\): The GP decreases in value (becomes more negative) but increases in absolute magnitude, diverging negatively as the sequence progresses.
4. \(a < 0, 0 < r < 1\): The GP approaches zero from the negative side, decreasing in absolute value, but remains negative throughout due to a negative starting term and a fractional common ratio.
5. \(a \neq 0, r < 0\): The GP oscillates, with terms alternating in sign. The sequence does not strictly increase or decrease but grows in absolute value if \(|r| > 1\) or diminishes in absolute value if \(|r| < 1\).

When \(r = 1\) in a geometric progression (GP), each term of the sequence is equal to the first term \(a\), resulting in a constant sequence. This means that the GP does not change from term to term, remaining the same throughout the sequence. For any \(n\), the \(n\)-th term \(a_n\) is always equal to \(a\), illustrating a special case of a geometric progression where the common ratio does not lead to growth or decay in the sequence. Thus, the series would be \(a, a, a, \ldots\), indefinitely.

Finite Geometric Series

A finite geometric series is a sum of terms in a geometric progression where each term is obtained by multiplying the previous term by a constant ratio, and the series has a specific number of terms. Formally, if \(a\) is the first term, \(r\) the common ratio, and \(n\) the total number of terms, the finite geometric series is given by:

\[S_n = a + ar + ar^2 + \ldots + ar^{n-1}\]

The sum of a finite geometric series can be calculated using the formula:

\[S_n = \frac{a(1 - r^n)}{1 - r}\]

for \(r \neq 1\). This formula encapsulates the sum of all terms in the series, efficiently allowing for the computation of the series' total based on its initial term, common ratio, and length.

Proof:

To prove the formula for the sum of a finite geometric series of \(n\) terms, consider the series:

\[S_n = a + ar + ar^2 + ar^3 + \ldots + ar^{n-1}\]

Step 1: Multiply both sides by \(r\):

\[rS_n = ar + ar^2 + ar^3 + \ldots + ar^{n-1} + ar^n\]

Step 2: Shift the above series to the right by one term and subtract \(rS_n\) from \(S_n\):

\[ \begin{align*} S_n &= a + \phantom{a}\cancel{ar} + \cancel{ar^2} + \cancel{ar^3} + \ldots + \cancel{ar^{n-1}} \\ rS_n &= \phantom{a + {}} \cancel{ar} + \cancel{ar^2} + \cancel{ar^3} + \ldots + \cancel{ar^{n-1}} + ar^n \end{align*} \]

Subtracting the second equation from the first:

\[ \begin{align*} S_n - rS_n &= a - ar^n \\ S_n(1 - r) &= a(1 - r^n) \end{align*} \]

Step 3: Solve for \(S_n\):

If \(r \neq 1\), to find \(S_n\), divide both sides by \((1 - r)\):

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

This yields the formula for the sum of a finite geometric series of \(n\) terms, providing a concise method to calculate the total sum based on the first term \(a\), the common ratio \(r\), and the number of terms \(n\).

Examples

Example 1: Sum of the Geometric Series \(2 + 2\cdot3 + 2\cdot3^2 + \ldots + 2\cdot3^{10}\)

Solution: Identifying the Common Ratio:
The common ratio \(r\), which is the factor by which we multiply each term to get the next term, is correctly identified as \(3\).

Calculating the Sum:
To find the sum of the first \(n\) terms in this GP, we use the formula:

\[S_n = \frac{a(1 - r^n)}{1 - r}\]

Given \(n = 10\), \(a = 2\), and \(r = 3\), we substitute these into the formula:

\[S_n = \frac{2(1 - 3^{10})}{1 - 3}\]

Simplifying the expression:

\[S_n = \frac{2(1 - 3^{10})}{-2} = -(1 - 3^{10})\]

\[S_n = 3^{10} - 1\]

This formula accurately calculates the sum of the series \(2 + 2\cdot3 + 2\cdot3^2 + \ldots + 2\cdot3^{10}\) based on the corrected common ratio.

Example 2: Sum of the Geometric Series \((1/2) + (1/2)^3 + (1/2)^5 + \ldots\) up to 20 terms

Solution:

Identifying the Common Ratio:
The first term \(a = 1/2\), and the second term is \((1/2)^3\). To find the common ratio (\(r\)), divide the second term by the first:

\[r = \frac{(1/2)^3}{(1/2)} = (1/2)^2 = 1/4\]

This confirms the series is a geometric progression with \(a = 1/2\) and \(r = 1/4\).

Calculating the Sum:
For a geometric series up to 20 terms, use the formula:

\[S_n = \frac{a(1 - r^n)}{1 - r}\]

Given \(n = 20\), \(a = 1/2\), and \(r = 1/4\), we substitute these values:

\[S_n = \frac{(1/2)(1 - (1/4)^{20})}{1 - 1/4}\]

\[S_n = \frac{(1/2)(1 - (1/4)^{20})}{3/4}\]

\[S_n = \frac{2}{3}(1 - (1/4)^{20})\]

This expression provides the sum of the series \((1/2) + (1/2)^3 + (1/2)^5 + \ldots\) up to 20 terms.

Usual Mistake

An usual mistake that students make in the beginning with sum of geometric series formula is putting wrong number of terms \(n\) in the formula. Consider the following example:

Example: Sum of the Series \((2^{21}/3^2) + (2^{20}/3^3) + (2^{19}/3^4) + \ldots + (2^3/3^{20})\)

Identifying the Number of Terms (\(n\)) in the Series:

The powers of 3 in the denominators start from \(3^2\) and go up to \(3^{20}\). This indicates that the total number of terms is \(20 - 2 = 18\), as the sequence begins at the second term and concludes at the twentieth term. Thus, \(n = 18\).

Identifying the Common Ratio (\(r\)):

To find the common ratio, divide the second term by the first term:

\[ \frac{2^{20}/3^3}{2^{21}/3^2} = \frac{2^{20}}{3^3} \cdot \frac{3^2}{2^{21}} = \frac{2^{-1}}{3} = \frac{1}{2 \cdot 3} = \frac{1}{6} \]

Hence, the common ratio \(r = \frac{1}{6}\).

Calculating the Sum (\(S_n\)):

With \(n = 18\), \(a = \frac{2^{21}}{3^2}\), and \(r = \frac{1}{6}\), apply the formula for the sum of a finite geometric series:

\[S_n = \frac{a(1 - r^n)}{1 - r}\]

Substituting the identified values into the formula:

\[S_n = \frac{\frac{2^{21}}{3^2}(1 - (\frac{1}{6})^{18})}{1 - \frac{1}{6}}\]

\[S_n = \frac{\frac{2^{21}}{9}(1 - (\frac{1}{6})^{18})}{\frac{5}{6}}\]

\[S_n = \frac{2^{21} \cdot 6}{9.5}(1 - (\frac{1}{6})^{18})\]

\[S_n = \frac{2^{22}}{15}\left[1 - \frac{1}{6^{18}}\right]\]

Example

Prove the identity:

\[x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \ldots + y^{n-1} = \frac{x^n - y^n}{x - y}\]

Proof:

Start by factoring \(x^{n-1}\) out of the left-hand side (LHS):

\[LHS = x^{n-1}\left(1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2 + \left(\frac{y}{x}\right)^3 + \ldots + \left(\frac{y}{x}\right)^{n-1}\right)\]

The expression inside the parentheses is a geometric progression (GP) with the first term \(a = 1\) and the common ratio \(r = \frac{y}{x}\). The sum of the first \(n\) terms of a GP is given by:

\[S_n = \frac{a(1 - r^n)}{1 - r}\]

Applying this formula to the GP inside the parentheses:

\[S_n = \frac{1 - \left(\frac{y}{x}\right)^n}{1 - \frac{y}{x}}\]

Simplifying the denominator and numerator:

\[S_n = \frac{1 - \frac{y^n}{x^n}}{1 - \frac{y}{x}} = \frac{x^n - y^n}{x^n(1 - \frac{y}{x})}\]

Further simplification of the denominator inside the fraction yields:

\[S_n = \frac{x^n - y^n}{x^n \cdot \frac{x - y}{x}} = \frac{x^n - y^n}{x - y}\]

Multiplying this result by \(x^{n-1}\) gives us the original left-hand side:

\[LHS = x^{n-1} \cdot \frac{x^n - y^n}{x - y}\]

Since the manipulation of the LHS leads directly to the right-hand side (RHS) of the original equation:

\[\frac{x^n - y^n}{x - y}\]

We have proven that:

\[x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \ldots + y^{n-1} = \frac{x^n - y^n}{x - y}\]

This identity elegantly demonstrates how a geometric progression can be used to simplify and prove polynomial identities, showcasing the interplay between algebraic expressions and summation formulas.

Very Important Identities

Using the example above we can derive a few very important identities that you will be seeing again and again in the future.

Factorization of \(x^n - y^n\) and \(x^n + y^n\)

1. For any \(n\), positive integer:

\[x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + x^{n-3}y^2 + \ldots + y^{n-1})\]

1. For \(n\) odd:

\[x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \ldots + y^{n-1})\]

Note: When \(n\) is even, \(x^n + y^n\) cannot be factored in this manner.

Proof of the Second Identity:

To prove the identity for \(x^n + y^n\) when \(n\) is odd, we utilize the first identity by substituting \(y\) with \(-y\):

\[x^n - (-y)^n = (x - (-y))(x^{n-1} + x^{n-2}(-y) + x^{n-3}(-y)^2 + \ldots + (-y)^{n-1})\]

Given \(n\) is odd, \((-y)^n = -y^n\), and the expression simplifies to:

\[x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \ldots + (-1)^{n-1}y^{n-1})\]

Since \(n\) is odd, the last term \(y^{n-1}\) remains positive due to the odd power of \(-1\), leading to the final form:

\[x^n + y^n = (x + y)(x^{n-1} - x^{n-2}y + x^{n-3}y^2 - \ldots + y^{n-1})\]

Factorization of \(x^n - y^n\) When \(n\) is Eeen

Let \(n = 2m\), where \(m\) is an integer. This representation indicates that \(n\) is even. We aim to show that for even \(n\), \(x^n - y^n\) can be factored into \((x - y)\) and \((x + y)\) among other factors.

Starting with \(x^n - y^n\) and substituting \(n = 2m\), we have:

\[x^n - y^n = x^{2m} - y^{2m}\]

This can be expressed as a difference of squares since \(x^{2m} = (x^2)^m\) and \(y^{2m} = (y^2)^m\):

\[x^{2m} - y^{2m} = (x^2)^m - (y^2)^m\]

\[(x^2)^m - (y^2)^m = (x^2 - y^2)((x^2)^{m-1} + (x^2)^{m-2}(y^2) + \ldots + (y^2)^{m-1})\]

Since \(x^2 - y^2\) is itself a difference of squares, it further factors into:

\[(x^2 - y^2) = (x - y)(x + y)\]

Thus, combining these results, we obtain:

\[x^{2m} - y^{2m} = (x - y)(x + y)((x^2)^{m-1} + (x^2)^{m-2}(y^2) + \ldots + (y^2)^{m-1})\]

Infinite Geometric Series and Its Sum

Definition of an Infinite Geometric Series

An infinite geometric series is defined as the sum of infinitely many terms of the form:

\[ S = a + ar + ar^2 + ar^3 + \cdots \]

where \(a\) is the first term and \(r\) is the common ratio, with \(a \neq 0\) and \(r \neq 0\).

Convergence Criteria

For an infinite geometric series to converge to a finite value, the absolute value of the common ratio must be less than one (\(|r| < 1\)). If \(|r| \geq 1\), the series diverges, meaning it grows without bound.

Sum for the Infinite Series

Finite Geometric Series

Before addressing the infinite case, let's consider a finite geometric series with \(n\) terms:

\[ S_n = a + ar + ar^2 + \cdots + ar^{n-1} \]

We know that the sum of this series can be found using the formula:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Transition to Infinity

To examine the behavior of \(S_n\) as \(n\) approaches infinity, we rewrite the formula for \(S_n\) by dividing the numerator and the denominator by \(1-r\), yielding:

\[ S_n = \frac{a}{1-r} - \frac{ar^n}{1-r} \]

As \(n\) becomes larger and approaches infinity, and given \(|r| < 1\), the term \(r^n\) approaches zero. This behavior is due to the property of geometric progressions where, if the absolute value of the ratio is less than one, the terms of the sequence become closer and closer to zero as \(n\) increases.

With the understanding that \(r^n \rightarrow 0\) as \(n \rightarrow \infty\) for \(|r| < 1\), we can now consider the limit of \(S_n\) as \(n\) approaches infinity:

\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{a}{1-r} - \frac{ar^n}{1-r} \right) \]

Given \(r^n \rightarrow 0\), the second term in the expression above vanishes, leading to:

\[ S_{\infty} = \frac{a}{1-r} \]

This result shows that the sum of an infinite geometric series converges to \(a/(1-r)\) under the condition that \(|r| < 1\). The convergence to a finite sum highlights a remarkable property of infinite geometric series, where despite the addition of infinitely many terms, the series sums to a finite, well-defined value.

Example

Problem:

Find the sum of the infinite geometric series: \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \).

Solution:

Given the series:

\[ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \]

The first term \(a\) is \(\frac{1}{2}\), and the common ratio \(r\) is also \(\frac{1}{2}\).

The sum \(S_{\infty}\) of an infinite geometric series is given by:

\[ S_{\infty} = \frac{a}{1 - r} \]

Substituting the given values:

\[ S_{\infty} = \frac{\frac{1}{2}}{1 - \frac{1}{2}} \]

Simplifying:

\[ S_{\infty} = 1 \]

Conclusion:

The sum of the series \( \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots \) up to infinity is \(1\).

Example

Problem:

Prove that for \( |x| < 1 \), the expression \( (1 + x)^{-1} \) equals the infinite series \( 1 + x + x^2 + x^3 + \cdots \).

Proof:

Given \( |x| < 1 \), we aim to prove that \( (1 + x)^{-1} = 1 + x + x^2 + x^3 + \cdots \).

Consider the infinite geometric series \( S = 1 + x + x^2 + x^3 + \cdots \) with the first term \( a = 1 \) and common ratio \( r = x \). For the series to converge, it is necessary that \( |r| < 1 \), which is satisfied by our condition \( |x| < 1 \).

The sum of an infinite geometric series is given by \( S_{\infty} = \frac{a}{1 - r} \). Substituting \( a = 1 \) and \( r = x \) into this formula, we get:

\[ S_{\infty} = \frac{1}{1 - x} \]

This expression is equivalent to \( (1 + x)^{-1} \) by basic algebraic principles, under the assumption that \( |x| < 1 \).

Hence, \( (1 + x)^{-1} = 1 + x + x^2 + x^3 + \cdots \) for \( |x| < 1 \), proving the given statement.

Important properties of GP

Three terms in GP

For three numbers \(a\), \(b\), and \(c\) in geometric progression (GP), the relationship \(b^2 = ac\) holds.

Proof:

By definition, in a geometric progression, each term after the first is obtained by multiplying the preceding term by a constant ratio, denoted as \(r\). Given \(a\), \(b\), and \(c\) are consecutive terms of a GP, they satisfy:

\[ b = ar \quad \text{and} \quad c = br \]

From these relations, the common ratio \(r\) can be expressed as \(r = \frac{b}{a}\). Substituting this expression into \(c = br\), we get:

\[ c = b \left( \frac{b}{a} \right) = \frac{b^2}{a} \]

Multiplying both sides by \(a\) yields:

\[ ac = b^2 \]

This completes the proof, establishing that for three numbers \(a\), \(b\), and \(c\) in geometric progression, the property \(b^2 = ac\) indeed holds.

Also, observe that \(abc=b^3\)

kth term from the end in GP

In a geometric progression (GP) consisting of terms \(a, ar, ar^2, ar^3, \ldots, ar^{n-1}\),

where \(a\) is the first term,

\(r\) is the common ratio, and \(n\) is the total number of terms, athe \(k^{th}\) term from the end is given by \(ar^{n-k}\).

Proof:

Consider a GP defined by the sequence \(a, ar, ar^2, ar^3, \ldots, ar^{n-1}\), where:

• \(a\) is the first term,
• \(r\) is the common ratio,
• \(n\) represents the total number of terms.

The general formula for the \(m^{th}\) term (\(T_m\)) of a GP is given by:

\[ T_m = ar^{m-1} \]

To find the \(k^{th}\) term from the end, we note that the last term (\(T_n\)) is \(ar^{n-1}\). Counting backwards \(k\) terms from the end essentially means finding the \((n-k+1)^{th}\) term from the beginning, since the last term is the \(n^{th}\) term, the second-to-last is the \((n-1)^{th}\) term, and so on, until we reach the \((n-k+1)^{th}\) term from the beginning.

Applying the general formula for the \(m^{th}\) term to find the \((n-k+1)^{th}\) term, we substitute \(m = n-k+1\) into \(T_m = ar^{m-1}\), yielding:

\[ T_{n-k+1} = ar^{(n-k+1)-1} = ar^{n-k} \]

Thus, the \(k^{th}\) term from the end of the GP \(a, ar, ar^2, ar^3, \ldots, ar^{n-1}\) is \(ar^{n-k}\), proving the property.

Constant Product of Equidistant Terms in a Geometric Progression

In a geometric progression (GP) consisting of \(n\) terms, represented as \(a_1, a_2, a_3, \ldots, a_n\), with \(a_1\) being the first term and \(r\) the common ratio, the product of terms equidistant from the start and end of the series is constant across the series. For any term \(a_k\) and its counterpart \(a_{n-k+1}\) equidistant from the opposite end, the product is given by:

\[ a_1 \cdot a_n = a_2 \cdot a_{n-1} = \cdots = a_k \cdot a_{n-k+1} = \text{constant} = a_1^2 \cdot r^{n-1} \]

Proof:

Given a GP with the first term \(a_1 = a\) and common ratio \(r\), the \(k^{th}\) term (\(a_k\)) and the \((n-k+1)^{th}\) term (\(a_{n-k+1}\)) can be expressed as:

\[ a_k = a \cdot r^{k-1} \]

\[ a_{n-k+1} = a \cdot r^{(n-k+1)-1} = a \cdot r^{n-k} \]

The product of these two terms is:

\[ a_k \cdot a_{n-k+1} = (a \cdot r^{k-1}) \cdot (a \cdot r^{n-k}) \]

\[ = a^2 \cdot r^{(k-1)+(n-k)} \]

\[ = a^2 \cdot r^{n-1} \]

This demonstrates that the product of the terms equidistant from the start and end of the GP (\(a_k \cdot a_{n-k+1}\)) is indeed constant and equals \(a^2 \cdot r^{n-1}\) for any \(k\) within the range \(1 \leq k \leq n\). This constant nature of the product underlines a fundamental symmetry in geometric progressions, highlighting that the product of terms equidistant from the ends is invariant, regardless of the position \(k\) within the sequence.

Product of all terms of a GP

For a geometric progression (GP) with \(n\) terms, denoted by \(a, ar, ar^2, ar^3, \ldots, ar^{n-1}\), where \(a\) is the first term and \(r\) is the common ratio, the product of all terms in the sequence is given by \(a^n r^{\frac{n(n-1)}{2}}\).

Proof:

Consider a GP defined by the terms \(a, ar, ar^2, ar^3, \ldots, ar^{n-1}\). The product \(P\) of all terms in the GP can be written as:

\[ P = a \cdot ar \cdot ar^2 \cdot ar^3 \cdot \ldots \cdot ar^{n-1} \]

This expression simplifies to:

\[ P = a^n \cdot r^{1+2+3+\ldots+(n-1)} \]

The exponent of \(r\) is the sum of the first \(n-1\) natural numbers, which can be calculated using the formula for the sum of an arithmetic series:

\[ S = \frac{n(n-1)}{2} \]

Therefore, the product \(P\) can be expressed as:

\[ P = a^n \cdot r^{\frac{n(n-1)}{2}} \]

This demonstrates that the product of all terms in a GP is \(a^n\) multiplied by \(r\) raised to the sum of the first \(n-1\) natural numbers, which is equivalent to \(a^n r^{\frac{n(n-1)}{2}}\). This result highlights the relationship between the geometric progression's parameters and the product of its terms, encapsulating the progression's multiplicative nature in a concise formula.

Arithmetic Operations on a GP

1. Multiplication by a Real Number \(k\)

   When each term of a GP with a common ratio \(r\) is multiplied by a real number \(k\) (where \(k\) is not equal to zero), the resulting sequence remains a GP. The common ratio of the new GP is unchanged at \(r\).

   Proof: Consider a GP \(\{a, ar, ar^2, \ldots\}\). Multiplying each term by \(k\) results in the sequence \(\{ka, kar, kar^2, \ldots\}\), which is still geometric with the same common ratio \(r\), as \(\frac{kar}{ka} = r\).

2. Division by \(k\)

   Similarly, dividing each term of a GP by a non-zero real number \(k\) results in a sequence that is also a GP, with the common ratio remaining \(r\).

   Proof: Starting with a GP \(\{a, ar, ar^2, \ldots\}\) and dividing each term by \(k\), we obtain \(\{\frac{a}{k}, \frac{ar}{k}, \frac{ar^2}{k}, \ldots\}\). This sequence maintains the common ratio \(r\), as \(\frac{\frac{ar}{k}}{\frac{a}{k}} = r\).

3. Addition or Subtraction by \(k\)

   Adding or subtracting a real number \(k\) (where \(k\) is not equal to zero) to each term of a GP does not result in a GP.

   Proof: Consider a GP \(\{a, ar, ar^2, \ldots\}\). Adding \(k\) to each term results in \(\{a+k, ar+k, ar^2+k, \ldots\}\). The ratio between successive terms varies, as \(\frac{ar+k}{a+k} \neq \frac{ar^2+k}{ar+k}\) in general, hence the resulting sequence is not a GP.

4. Logarithmic Transformation and Exponentiation

   • Logarithmic Transformation: Taking the logarithm base \(b\) of each term in a GP \(\{a, ar, ar^2, \ldots\}\) results in an arithmetic progression (AP) with a common difference of \(\log_b r\).

     Proof: The logarithm base \(b\) of the GP terms gives \(\{\log_b a, \log_b ar, \log_b ar^2, \ldots\}\), which simplifies to \(\{\log_b a, \log_b a + \log_b r, \log_b a + 2\log_b r, \ldots\}\), clearly an AP with common difference \(\log_b r\).

   • Exponentiation: Conversely, exponentiating each term of an AP by a base \(b\) results in a GP.

     Proof: Consider an AP \(\{a, a+d, a+2d, \ldots\}\). Exponentiating each term by base \(b\) yields \(\{b^a, b^{a+d}, b^{a+2d}, \ldots\}\). This forms a GP, as the ratio between successive terms is \(b^d\), showing exponential growth by a constant factor. 5. Raising Each Term of a Geometric Progression to a Power \(k\)

   Given a geometric progression (GP) with terms \(a, ar, ar^2, \ldots\) and a common ratio \(r\), when each term of the GP is raised to a certain power \(k\), the resulting sequence \(\{a^k, (ar)^k, (ar^2)^k, \ldots\}\) is also a geometric progression with a common ratio of \(r^k\).

5. Combining Terms of Two Geometric Progressions via Arithmetic Operations

   Given two geometric progressions (GPs) \(G_1\) and \(G_2\) with terms \(\{a_1, a_1r_1, a_1r_1^2, \ldots\}\) and \(\{a_2, a_2r_2, a_2r_2^2, \ldots\}\) respectively, where \(a_1\) and \(a_2\) are the first terms and \(r_1\) and \(r_2\) are the common ratios, the result of combining corresponding terms of \(G_1\) and \(G_2\) via arithmetic operations (multiplication, division, addition, subtraction) follows specific patterns:

   1. Multiplication: Produces a GP with the first term \(a_1a_2\) and common ratio \(r_1r_2\).
   2. Division: Results in a sequence that is a GP with the first term \(a_1/a_2\) (assuming \(a_2 \neq 0\)) and common ratio \(r_1/r_2\) (assuming \(r_2 \neq 0\)).
   3. Addition or Subtraction: Does not generally result in a GP.

   Proof:

   Multiplication:

   For the multiplication operation, combining corresponding terms of \(G_1\) and \(G_2\) gives:

   \[ \{a_1a_2, (a_1r_1)(a_2r_2), (a_1r_1^2)(a_2r_2^2), \ldots\} \]

   This sequence is a GP because the ratio of any term to its preceding term is constant:

   \[ \frac{(a_1r_1^n)(a_2r_2^n)}{(a_1r_1^{n-1})(a_2r_2^{n-1})} = r_1r_2 \]

   Division:

   For division, assuming \(a_2\) and \(r_2\) are not zero, the sequence formed by combining corresponding terms is:

   \[ \left\{\frac{a_1}{a_2}, \frac{a_1r_1}{a_2r_2}, \frac{a_1r_1^2}{a_2r_2^2}, \ldots\right\} \]

   This is also a GP, with each term's ratio to its preceding term being \(r_1/r_2\).

   Addition and Subtraction:

   Addition or subtraction of corresponding terms yields:

   \[ \{a_1 + a_2, a_1r_1 + a_2r_2, a_1r_1^2 + a_2r_2^2, \ldots\} \]

   \[ \{a_1 - a_2, a_1r_1 - a_2r_2, a_1r_1^2 - a_2r_2^2, \ldots\} \]

   The sequences resulting from addition or subtraction do not generally form GPs because the difference (or sum) between successive terms does not multiply by a constant ratio. The ratio of successive terms varies, depending on the values of \(r_1\), \(r_2\), \(a_1\), and \(a_2\).

6. Forming a Subsequence from a Geometric Progression by Selecting Terms at a Constant Interval

   Given a geometric progression (GP) with a common ratio \(r\), if we form a subsequence by choosing terms at a constant interval of length \(k\), the resulting subsequence is also a GP with a common ratio of \(r^k\).

   Definition of Interval

   In the context of sequences, an interval of length \(k\) refers to the practice of selecting every \(k^{th}\) term from a sequence, effectively skipping \(k-1\) terms between selections. For example, an interval of length \(k=2\) means selecting every second term from the sequence.

   Proof:

   Consider a GP given by the sequence \(\{a, ar, ar^2, ar^3, \ldots\}\), where \(a\) is the first term and \(r\) is the common ratio.

   Original GP:

   • The \(n^{th}\) term of this GP can be represented as \(ar^{n-1}\).

   Forming a Subsequence:

   • By selecting terms at a constant interval of length \(k\), we choose the terms \(a, ar^k, ar^{2k}, ar^{3k}, \ldots\). This selection corresponds to taking every \(k^{th}\) term from the original sequence.

   Analysis of the Subsequence:

   • The first term of the subsequence is \(a\).
   • The second term is \(ar^k\).
   • The third term is \(ar^{2k}\), and so on.

   Determining the Common Ratio:

   • The ratio of the second term to the first term in the subsequence is \(\frac{ar^k}{a} = r^k\).
   • Similarly, the ratio of the third term to the second term is \(\frac{ar^{2k}}{ar^k} = r^k\).

   Given that the ratio between successive terms in the subsequence is constant and equal to \(r^k\), the subsequence formed by selecting every \(k^{th}\) term from the original GP is itself a GP with a common ratio of \(r^k\).

Recurring decimals and GP

Definition of Recurring Decimals

A recurring decimal is a decimal number in which a digit or group of digits repeats infinitely. For example, in \(0.6666\ldots\), the digit \(6\) repeats indefinitely, and in \(0.123123\ldots\), the sequence \(123\) repeats.

Recurring Decimals and Infinite Geometric Progressions

Recurring decimals can be represented as infinite geometric progressions (GPs). This representation is based on the observation that each repetition of the recurring digits can be seen as a term in a GP, where the common ratio is a power of \(10\) inversely proportional to the length of the repeating sequence.

Example: Converting a Recurring Decimal to an Infinite GP

Consider the recurring decimal \(0.232323\ldots\). This can be expressed as the sum of an infinite GP:

\[ 0.23 + 0.0023 + 0.000023 + \ldots \]

In this series: - The first term (\(a\)) is \(0.23\) or \(23 \times 10^{-2}\). - The common ratio (\(r\)) is \(10^{-2}\), since each subsequent term is \(100\) times smaller than the previous term.

The series can be rewritten as:

\[ 23 \times 10^{-2} + 23 \times 10^{-4} + 23 \times 10^{-6} + \ldots \]

Summing the Infinite GP

The sum (\(S\)) of an infinite GP is given by the formula \(S = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio. Applying this formula to our series:

\[ S = \frac{23 \times 10^{-2}}{1 - 10^{-2}} \]

Calculating the sum:

\[ S = \frac{23 \times 10^{-2}}{1 - 0.01} \]

\[ S = \frac{23 \times 10^{-2}}{0.99} \]

\[ S = \frac{23}{99} \]

Conclusion

The recurring decimal \(0.232323\ldots\) can be expressed as the infinite geometric progression \(23 \times 10^{-2} + 23 \times 10^{-4} + 23 \times 10^{-6} + \ldots\), which sums to \(\frac{23}{99}\). This example demonstrates how recurring decimals are not just patterns in decimal notation but can also be precisely represented and summed using the concept of infinite GPs. This approach provides a powerful tool for converting recurring decimals into exact fractional forms.

Example

Consider the recurring decimal \(3.142857142857\ldots\). This decimal has a repeating cycle of \(142857\) that goes on infinitely. To express this recurring decimal as an infinite geometric progression, we first isolate the repeating part:

\[ 3.142857142857\ldots = 3 + 0.142857 + 0.000000142857 + 0.000000000142857 + \ldots \]

The sequence \(0.142857, 0.000000142857, 0.000000000142857, \ldots\) forms an infinite GP, where: - The first term (\(a\)) is \(0.142857\), or more precisely, \(142857 \times 10^{-6}\). - The common ratio (\(r\)) is \(10^{-6}\), since each subsequent term is \(10^6\) times smaller than the previous one, correcting the earlier error.

The GP can thus be written as:

\[ 142857 \times 10^{-6} + 142857 \times 10^{-12} + 142857 \times 10^{-18} + \ldots \]

Summing the Infinite GP

The sum \(S\) of an infinite GP is given by the formula \(S = \frac{a}{1 - r}\), substituting the values:

\[ S = \frac{142857 \times 10^{-6}}{1 - 10^{-6}} \]

Simplifying, we get:

\[ S = \frac{142857 \times 10^{-6}}{0.999999} \]

\[ S = \frac{142857}{999999} \]

Including the non-repeating part (\(3\)) yields the total value of the recurring decimal:

\[ 3 + \frac{142857}{999999} \]

which can be simlified as \(3 + \frac{142857}{999999} = \frac{22}{7}\)