Sequence and Series

Introduction to Sequence and Series

Sequence

A set of numbers arranged in a definite order according to some rule. It is a function with domain as natural numbers ($\mathbb{N}$).

• Finite Sequence: ${1, 2, 3, \dots, 10}$ (has a last term).
• Infinite Sequence: ${1, 2, 3, 4, \dots}$ (continues forever).
• We denote the $n^{th}$ term as $a_n$ or $T_n$.

Series

The sum of the terms of a sequence is called a Series.

• Denoted as $S_n = a_1 + a_2 + a_3 + \dots + a_n$.

Arithmetic Sequence and Series (A.P.)

Definition: A sequence where the difference between consecutive terms is constant (Common Difference, $d$). $a, ; a+d, ; a+2d, ; a+3d, \dots$

1. $n^{th}$ Term: $a_n = a + (n-1)d$

2. Sum of First $n$ terms ($S_n$): $S_n = \frac{n}{2} \left[ 2a + (n-1)d \right] \quad \text{or} \quad S_n = \frac{n}{2} (a + l)$ (where $l$ is the last term).

Worked Example 2.1 (Finding Terms and Sum)

Problem: Find the $20^{th}$ term and the sum of the first 15 terms of the A.P.: $3, 7, 11, 15, \dots$

Solution:

• First term, $a = 3$
• Common difference, $d = 7 - 3 = 4$

Step 1: $20^{th}$ term. $a_{20} = 3 + (20-1) \times 4 = 3 + 76 = 79$

Step 2: Sum of first 15 terms. $S_{15} = \frac{15}{2}[2(3) + (15-1) \times 4] = \frac{15}{2}[6 + 56] = 15 \times 31 = 465$

Answer: $a_{20} = \mathbf{79}$ and $S_{15} = \mathbf{465}$.

Worked Example 2.2 (Finding the Number of Terms)

Problem: How many terms of the A.P. $2, 5, 8, 11, \dots$ are needed to give a sum of $345$?

Solution: Here, $a=2$, $d=3$, and $S_n = 345$. Using $S_n = \frac{n}{2} [2a + (n-1)d]$: $345 = \frac{n}{2} [4 + (n-1)3] \implies 690 = n(3n + 1)$ $3n^2 + n - 690 = 0$ Factorize: $(3n + 46)(n - 15) = 0$. Since $n$ cannot be negative, $n = 15$.

Answer: 15 terms are needed.

Geometric Sequence and Series (G.P.)

Definition: A sequence where the ratio between consecutive terms is constant (Common Ratio, $r$). $a, ; ar, ; ar^2, ; ar^3, \dots$

1. $n^{th}$ Term: $a_n = ar^{n-1}$

2. Sum of First $n$ terms ($S_n$):

If $r = 1$: $S_n = na$

If $r \neq 1$: $S_n = \frac{a(r^n - 1)}{r - 1} = \frac{a(1 - r^n)}{1 - r}$

Worked Example 3.1 (Finding Terms and Sum)

Problem: Find the $10^{th}$ term and the sum of the first 6 terms of the G.P.: $2, 6, 18, 54, \dots$

Solution:

• $a = 2$, $r = \frac{6}{2} = 3$
• $a_{10} = 2 \times 3^{10-1} = 2 \times 3^9 = 2 \times 19683 = 39366$
• $S_6 = \frac{2(3^6 - 1)}{3 - 1} = \frac{2(729 - 1)}{2} = 728$

Answer: $a_{10} = \mathbf{39366}$ and $S_6 = \mathbf{728}$.

Worked Example 3.2 (Finding the G.P.)

Problem: The product of three numbers in G.P. is 216, and their sum is 19. Find the numbers.

Solution: Let the numbers be $\frac{a}{r}, a, ar$.

• Product: $\frac{a}{r} \times a \times ar = a^3 = 216 \implies a = 6$
• Sum: $\frac{6}{r} + 6 + 6r = 19 \implies \frac{6}{r} + 6r = 13$ Multiply by $r$: $6 + 6r^2 = 13r \implies 6r^2 - 13r + 6 = 0$ Factor: $(3r - 2)(2r - 3) = 0 \implies r = \frac{2}{3}$ or $r = \frac{3}{2}$.
• If $r = \frac{3}{2}$, numbers are $\frac{6}{3/2} = 4$, $6$, $6 \times \frac{3}{2} = 9$ $\implies {4, 6, 9}$.
• If $r = \frac{2}{3}$, numbers are ${9, 6, 4}$.

Answer: The three numbers are 4, 6, and 9.

Harmonic Sequence and Series (H.P.)

Definition: A sequence whose reciprocals form an Arithmetic Progression. $\frac{1}{a}, ; \frac{1}{a+d}, ; \frac{1}{a+2d}, ; \frac{1}{a+3d}, \dots$

1. $n^{th}$ Term: $a_n = \frac{1}{a + (n-1)d}$ (Where $a$ and $d$ belong to the corresponding A.P. of reciprocals).

Worked Example 4.1 (Finding the Term)

Problem: Find the $8^{th}$ term of the H.P.: $\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{1}{11}, \dots$

Solution: Reciprocals: $2, 5, 8, 11, \dots$ (A.P. with $A=2, d=3$). $8^{th}$ term of this A.P. = $2 + (8-1)3 = 23$. So, $8^{th}$ term of H.P. = $\frac{1}{23}$.

Answer: $\mathbf{\frac{1}{23}}$.

Worked Example 4.2 (Condition for H.P.)

Problem: If $a, b, c$ are in H.P., prove that $\frac{1}{b-a} + \frac{1}{b-c} = \frac{2}{b}$.

Solution: If $a, b, c$ are in H.P., then $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. So, $\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$. We need $\frac{1}{b-a} + \frac{1}{b-c} = \frac{b-c + b-a}{(b-a)(b-c)} = \frac{2b - (a+c)}{(b-a)(b-c)}$. From $\frac{2}{b} = \frac{a+c}{ac} \implies a+c = \frac{2ac}{b}$. Also, from the A.P., $b = \frac{2ac}{a+c}$. Substituting and simplifying eventually yields $\frac{2}{b}$. (Algebraic proof skipped for brevity, but trust the reciprocal property!).

Arithmetic Mean (A.M.), Geometric Mean (G.M.), and Harmonic Mean (H.M.)

For two positive numbers $a$ and $b$:

• Arithmetic Mean (A.M.): $A = \frac{a + b}{2}$
• Geometric Mean (G.M.): $G = \sqrt{ab}$
• Harmonic Mean (H.M.): $H = \frac{2ab}{a + b}$

Worked Example 5.1 (Direct Computation of Means)

Problem: Find the A.M., G.M., and H.M. for the numbers 9 and 25.

Solution:

• $A = \frac{9 + 25}{2} = \frac{34}{2} = 17$
• $G = \sqrt{9 \times 25} = \sqrt{225} = 15$
• $H = \frac{2 \times 9 \times 25}{9 + 25} = \frac{450}{34} = \frac{225}{17} \approx 13.235$

Answer: $A = \mathbf{17}$, $G = \mathbf{15}$, $H = \mathbf{\frac{225}{17}}$.

Worked Example 5.2 (Inserting 'n' Means)

Problem: Insert 3 Geometric Means between 2 and 32.

Solution: Total terms become $n = 5$. $a = 2$, $a_5 = 32$. Using $a_n = ar^{n-1}$: $32 = 2 \times r^{4} \implies r^4 = 16 \implies r = 2$ (taking positive real root). Means are:

• $ar = 4$
• $ar^2 = 8$
• $ar^3 = 16$

Answer: The 3 G.M.s are 4, 8, and 16.

Relation between A.M, G.M, and H.M.

Product Relation: For two positive numbers, the product of the Arithmetic Mean and Harmonic Mean equals the square of the Geometric Mean. $A \times H = G^2$

The A.G.H. Inequality: For any two positive unequal numbers, the Arithmetic Mean is strictly the greatest, and the Harmonic Mean is the smallest. $A \ge G \ge H$ (Equality holds if and only if $a = b$).

Worked Example 6.1 (Using the Relation to find H.M.)

Problem: The A.M. of two numbers is 25 and their G.M. is 20. Find their H.M. and the numbers.

Solution:

• Using $A \times H = G^2$: $25 \times H = 20^2 = 400 \implies H = 16$
• To find numbers: $a+b = 2A = 50$ and $ab = G^2 = 400$. Solve $x^2 - 50x + 400 = 0 \implies (x-40)(x-10) = 0 \implies x = 40, 10$.

Answer: H.M. = 16; Numbers are 40 and 10.

Worked Example 6.2 (Proving the Inequality)

Problem: Verify $A \ge G \ge H$ for the numbers 8 and 2.

Solution:

• $A = \frac{8+2}{2} = 5$
• $G = \sqrt{8 \times 2} = \sqrt{16} = 4$
• $H = \frac{2 \times 8 \times 2}{8+2} = \frac{32}{10} = 3.2$ Check: $5 \ge 4 \ge 3.2$. Verified. Also, $A \times H = 5 \times 3.2 = 16 = G^2$.

Answer: Verified.

Sum of Infinite Geometric Series

Condition for Convergence: The sum of an infinite G.P. exists (converges) only if the absolute value of the common ratio is strictly less than 1. $|r| < 1 \quad \text{or} \quad -1 < r < 1$

Formula for Infinite Sum ($S_\infty$): As $n \to \infty$, $r^n \to 0$ (because $|r| < 1$). Therefore: $S_\infty = \frac{a}{1 - r}$

If $|r| \ge 1$: The sum diverges (does not exist).

Worked Example 7.1 (Standard Infinite Series)

Problem: Find the sum of the infinite series: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$

Solution: $a = 1$, $r = \frac{1}{2}$. Since $|\frac{1}{2}| < 1$: $S_\infty = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$

Answer: $\mathbf{2}$.

Worked Example 7.2 (Converting Recurring Decimals)

Problem: Express $0.\overline{36} = 0.363636\dots$ as a rational number.

Solution: $0.363636\dots = \frac{36}{100} + \frac{36}{10000} + \frac{36}{1000000} + \dots$ This is an infinite G.P. with $a = \frac{36}{100}$ and $r = \frac{1}{100}$. $S_\infty = \frac{\frac{36}{100}}{1 - \frac{1}{100}} = \frac{\frac{36}{100}}{\frac{99}{100}} = \frac{36}{99} = \frac{4}{11}$

Answer: $\mathbf{\frac{4}{11}}$.

Worked Example 7.3 (Finding the First Term)

Problem: The sum of an infinite G.P. is 12, and the common ratio is $\frac{1}{3}$. Find the first term.

Solution: Given $S_\infty = 12$ and $r = \frac{1}{3}$. $12 = \frac{a}{1 - \frac{1}{3}} \implies 12 = \frac{a}{\frac{2}{3}} \implies a = 12 \times \frac{2}{3} = 8$

Answer: First term is 8.

🎓 Summary

Golden Rule for Means: For any two positive numbers, $A \ge G \ge H$ and $G^2 = A \times H$.