Arithmetic Progression and Geometric Progression

In mathematics, sequences and series are fundamental concepts that appear frequently in various branches, including algebra, calculus, and discrete mathematics. They form the bedrock for understanding patterns, growth, and accumulation. Within this broad field, Arithmetic Progressions (AP) and Geometric Progressions (GP) are two of the most basic and widely studied types of sequences. Understanding their properties, formulas, and applications is crucial for solving many problems, especially in competitive examinations like JEE Main.

Arithmetic Progression (AP)

An Arithmetic Progression (AP) is a sequence of numbers such that the difference between consecutive terms is constant. This constant difference is called the common difference, usually denoted by 'd'.

Definition and Properties of AP

A sequence $a_1, a_2, a_3, \dots, a_n$ is an AP if for all $i \ge 1$, $a_{i+1} - a_i = d$, where 'd' is a constant.

The terms of an AP can be represented as: $a, a+d, a+2d, a+3d, \dots$ Here, 'a' is the first term of the sequence.

General Term (nth term) of an AP

The general term or the nth term of an AP, denoted by $a_n$, can be found using the formula: $a_n = a + (n-1)d$ where: $a_n$ is the nth term. $a$ is the first term. $n$ is the position of the term in the sequence. $d$ is the common difference.

Example: Consider the AP: 2, 5, 8, 11, ... Here, the first term $a = 2$. The common difference $d = 5 - 2 = 3$. To find the 5th term ($n=5$): $a_5 = a + (5-1)d = 2 + (4 \times 3) = 2 + 12 = 14$. The sequence is indeed 2, 5, 8, 11, 14, ...

Sum of the First n Terms of an AP

The sum of the first n terms of an AP, denoted by $S_n$, can be calculated using two formulas:

1. When the last term is not known: $S_n = \frac{n}{2} [2a + (n-1)d]$
2. When the last term ($l = a_n$) is known: $S_n = \frac{n}{2} [a + l]$

Example: For the AP: 2, 5, 8, 11, 14. Here, $n=5$, $a=2$, $d=3$, and the last term $l=14$. Using the first formula: $S_5 = \frac{5}{2} [2(2) + (5-1)3] = \frac{5}{2} [4 + (4 \times 3)] = \frac{5}{2} [4 + 12] = \frac{5}{2} [16] = 5 \times 8 = 40$. Using the second formula: $S_5 = \frac{5}{2} [2 + 14] = \frac{5}{2} [16] = 5 \times 8 = 40$. The sum of the terms is $2 + 5 + 8 + 11 + 14 = 40$.

Properties of AP

• If a constant 'k' is added to or subtracted from each term of an AP, the resulting sequence is also an AP with the same common difference.
• If each term of an AP is multiplied or divided by a non-zero constant 'k', the resulting sequence is also an AP with a common difference of $k \times d$ (or $d/k$).
• The sum of terms equidistant from the beginning and end of a finite AP is constant and equal to the sum of the first and last terms. For an AP $a_1, a_2, \dots, a_n$, $a_1 + a_n = a_2 + a_{n-1} = a_3 + a_{n-2} = \dots$.
• In an AP, any term is the arithmetic mean of the terms equidistant from it. For example, $a_k = \frac{a_{k-1} + a_{k+1}}{2}$.

Important Considerations for AP

When dealing with AP problems, it's often useful to represent three or four consecutive terms in a specific way to simplify calculations, especially when their sum or product is involved.

• Three consecutive terms: $a-d, a, a+d$
• Four consecutive terms: $a-3d, a-d, a+d, a+3d$ (common difference is 2d)

This choice of representation helps in cancelling out terms when summing them up.

Geometric Progression (GP)

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.

Definition and Properties of GP

A sequence $a_1, a_2, a_3, \dots, a_n$ is a GP if for all $i \ge 1$, $\frac{a_{i+1}}{a_i} = r$, where 'r' is a constant and $r \neq 0$. This constant is called the common ratio.

The terms of a GP can be represented as: $a, ar, ar^2, ar^3, \dots$ Here, 'a' is the first term of the sequence.

General Term (nth term) of a GP

The general term or the nth term of a GP, denoted by $a_n$, can be found using the formula: $a_n = a \cdot r^{n-1}$ where: $a_n$ is the nth term. $a$ is the first term. $n$ is the position of the term in the sequence. $r$ is the common ratio.

Example: Consider the GP: 3, 6, 12, 24, ... Here, the first term $a = 3$. The common ratio $r = \frac{6}{3} = 2$. To find the 5th term ($n=5$): $a_5 = a \cdot r^{5-1} = 3 \cdot 2^4 = 3 \cdot 16 = 48$. The sequence is 3, 6, 12, 24, 48, ...

Sum of the First n Terms of a GP

The sum of the first n terms of a GP, denoted by $S_n$, depends on the value of the common ratio 'r'.

1. When $r \neq 1$: $S_n = \frac{a(r^n - 1)}{r - 1}$ or $S_n = \frac{a(1 - r^n)}{1 - r}$
2. When $r = 1$: If $r=1$, all terms are the same ($a, a, a, \dots$). So, $S_n = n \cdot a$.

Example: For the GP: 3, 6, 12, 24, 48. Here, $n=5$, $a=3$, $r=2$. Using the formula for $r \neq 1$: $S_5 = \frac{3(2^5 - 1)}{2 - 1} = \frac{3(32 - 1)}{1} = 3(31) = 93$. The sum of the terms is $3 + 6 + 12 + 24 + 48 = 93$.

Sum to Infinity of a GP

If the absolute value of the common ratio $|r| < 1$, then the sum of an infinite geometric series converges to a finite value. This sum, denoted by $S_\infty$, is given by: $S_\infty = \frac{a}{1 - r}$, for $|r| < 1$. If $|r| \ge 1$, the sum to infinity does not converge (it diverges to infinity or oscillates).

Example: Consider the GP: 1, 1/2, 1/4, 1/8, ... Here, $a=1$ and $r = \frac{1/2}{1} = \frac{1}{2}$. Since $|r| = \frac{1}{2} < 1$, the sum to infinity exists. $S_\infty = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$. The sum of all terms in this infinite series is 2.

Properties of GP

• If each term of a GP is multiplied or divided by a non-zero constant 'k', the resulting sequence is also a GP with the same common ratio.
• If each term of a GP is added to or subtracted from a constant 'k', the resulting sequence is generally not a GP.
• The product of terms equidistant from the beginning and end of a finite GP is constant and equal to the product of the first and last terms. For a GP $a_1, a_2, \dots, a_n$, $a_1 \cdot a_n = a_2 \cdot a_{n-1} = a_3 \cdot a_{n-2} = \dots$.
• In a GP, the logarithm of each term forms an AP. That is, if $a_1, a_2, a_3, \dots$ is a GP, then $\log a_1, \log a_2, \log a_3, \dots$ is an AP.
• The common ratio of a GP can be found by dividing any term by its preceding term.

Important Considerations for GP

Similar to AP, representing consecutive terms in a specific way can simplify GP problems involving products or sums.

• Three consecutive terms: $\frac{a}{r}, a, ar$
• Four consecutive terms: $\frac{a}{r^3}, \frac{a}{r}, ar, ar^3$ (common ratio is $r^2$)
• Five consecutive terms: $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$

This representation is particularly useful when the product of the terms is given.

Relationship between AP and GP

Sometimes, problems involve sequences that are neither purely AP nor GP, but exhibit properties related to them.

• Arithmetic-Geometric Progression (AGP): A sequence where each term is the product of the corresponding terms of an AP and a GP. For example: $a, (a+d)r, (a+2d)r^2, (a+3d)r^3, \dots$. The sum of such a series requires a specific method involving multiplying by the common ratio 'r' and subtracting.
• Harmonic Progression (HP): A sequence whose reciprocals form an AP. If $a_1, a_2, a_3, \dots$ is an HP, then $\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \dots$ is an AP.

Example Problems and Solutions

Problem 1 (AP):

The sum of the first 10 terms of an AP is 120. The sum of the first 20 terms is 440. Find the first term and the common difference.

Solution:

Let the first term be 'a' and the common difference be 'd'. We are given $S_{10} = 120$ and $S_{20} = 440$. Using the formula $S_n = \frac{n}{2} [2a + (n-1)d]$: For $n=10$: $120 = \frac{10}{2} [2a + (10-1)d] \implies 120 = 5 [2a + 9d] \implies 24 = 2a + 9d$ (Equation 1). For $n=20$: $440 = \frac{20}{2} [2a + (20-1)d] \implies 440 = 10 [2a + 19d] \implies 44 = 2a + 19d$ (Equation 2). Now, we solve the system of linear equations: Subtract Equation 1 from Equation 2: $(2a + 19d) - (2a + 9d) = 44 - 24$ $10d = 20 \implies d = 2$. Substitute $d=2$ into Equation 1: $24 = 2a + 9(2) \implies 24 = 2a + 18 \implies 2a = 6 \implies a = 3$. So, the first term is 3 and the common difference is 2.

Problem 2 (GP):

Find the sum of the first 8 terms of the GP: 2, 6, 18, ...

Solution:

The first term is $a = 2$. The common ratio is $r = \frac{6}{2} = 3$. We need to find the sum of the first 8 terms ($n=8$). Using the formula $S_n = \frac{a(r^n - 1)}{r - 1}$ (since $r \neq 1$): $S_8 = \frac{2(3^8 - 1)}{3 - 1} = \frac{2(6561 - 1)}{2} = 6560$. The sum of the first 8 terms is 6560.

Problem 3 (GP - Sum to Infinity):

Find the sum to infinity of the GP: 8, -4, 2, -1, ...

Solution:

The first term is $a = 8$. The common ratio is $r = \frac{-4}{8} = -\frac{1}{2}$. Since $|r| = |-\frac{1}{2}| = \frac{1}{2} < 1$, the sum to infinity exists. Using the formula $S_\infty = \frac{a}{1 - r}$: $S_\infty = \frac{8}{1 - (-\frac{1}{2})} = \frac{8}{1 + \frac{1}{2}} = \frac{8}{\frac{3}{2}} = 8 \times \frac{2}{3} = \frac{16}{3}$. The sum to infinity is $\frac{16}{3}$.

Problem 4 (AP/GP Mixed Concept):

If $a, b, c$ are in AP and $a, b, c$ are also in GP, then which of the following is true? (A) $a=b=c$ (B) $a=b$ and $b \neq c$ (C) $a \neq b$ and $b=c$ (D) $a, b, c$ are not necessarily equal.

Solution:

If $a, b, c$ are in AP, then $b-a = c-b$, which implies $2b = a+c$. If $a, b, c$ are in GP, then $\frac{b}{a} = \frac{c}{b}$, which implies $b^2 = ac$. We have two conditions: 1) $2b = a+c$ 2) $b^2 = ac$ Substitute $b = \frac{a+c}{2}$ from (1) into (2): $(\frac{a+c}{2})^2 = ac$ $\frac{a^2 + 2ac + c^2}{4} = ac$ $a^2 + 2ac + c^2 = 4ac$ $a^2 - 2ac + c^2 = 0$ $(a-c)^2 = 0$ $a-c = 0 \implies a = c$. Now substitute $a=c$ back into $2b = a+c$: $2b = a+a \implies 2b = 2a \implies b = a$. Since $a=c$ and $b=a$, it means $a=b=c$. Therefore, the correct option is (A).