Chapter 8: Sequence and Series

8.1 Introduction

A sequence is an ordered list of numbers obeying a certain rule. Elements of a sequence are called its terms, usually denoted by \( a_1, a_2, a_3, \dots, a_n \). If a sequence has a finite number of terms, it is a finite sequence. Otherwise, it is an infinite sequence.

A series is the sum of elements of a sequence. If \( a_1, a_2, a_3, \dots, a_n \) limit the sequence, then the expression \( a_1 + a_2 + a_3 + \dots + a_n \) is called series.

8.2 Arithmetic Progression (A.P.)

A sequence \( a_1, a_2, \dots \) is an Arithmetic Progression (AP) if \( a_{n+1} - a_n = d \), a constant, for all \( n \ge 1 \). \( d \) is the common difference.

General Term: \( a_n = a + (n-1)d \)
Sum of first n terms: \( S_n = \frac{n}{2} [2a + (n-1)d] = \frac{n}{2} (a + l) \), where \( l \) is the last term.
Arithmetic Mean (A.M.): Given two numbers \( a \) and \( b \), their A.M. is \( A = \frac{a+b}{2} \). We can insert \( n \) numbers between \( a \) and \( b \) to form an A.P.

8.3 Geometric Progression (G.P.)

A sequence \( a_1, a_2, \dots \) with non-zero terms is a Geometric Progression (GP) if \( \frac{a_{k+1}}{a_k} = r \), a constant, for all \( k \ge 1 \). \( r \) is called the common ratio.

General Term: \( a_n = ar^{n-1} \)
Sum of first n terms: \[ S_n = \frac{a(r^n - 1)}{r - 1} \text{ if } r \neq 1 \] If \( r = 1 \), then \( S_n = na \).
Sum of an infinite G.P.: If \( |r| < 1 \), then sum up to infinity is: \[ S_\infty = \frac{a}{1 - r} \]
Geometric Mean (G.M.): Given two numbers \( a \) and \( b \), their G.M. is \( G = \sqrt{ab} \).

Relationship between A.M. and G.M.

Let \( A \) and \( G \) be A.M. and G.M. of two given positive real numbers \( a \) and \( b \), respectively. Then: \[ A = \frac{a+b}{2} \text{ and } G = \sqrt{ab} \] It can be shown that \( A \ge G \), because \[ A - G = \frac{a+b}{2} - \sqrt{ab} = \frac{a+b - 2\sqrt{ab}}{2} = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \ge 0 \]

Competency-Based Questions

Question 1:

The sum of the first \( p, q, r \) terms of an A.P. are \( a, b, c \), respectively. Prove that:

\[ \frac{a}{p}(q-r) + \frac{b}{q}(r-p) + \frac{c}{r}(p-q) = 0 \]

Answer 1:

Let the first term of the A.P. be \( A \) and common difference be \( D \).

Given:

\[ S_p = \frac{p}{2}[2A + (p-1)D] = a \Rightarrow \frac{a}{p} = A + \frac{1}{2}(p-1)D \]

\[ S_q = \frac{q}{2}[2A + (q-1)D] = b \Rightarrow \frac{b}{q} = A + \frac{1}{2}(q-1)D \]

\[ S_r = \frac{r}{2}[2A + (r-1)D] = c \Rightarrow \frac{c}{r} = A + \frac{1}{2}(r-1)D \]

Substitute these values into LHS of the given expression:

\[ LHS = \left[ A + (p-1)\frac{D}{2} \right] (q-r) + \left[ A + (q-1)\frac{D}{2} \right] (r-p) + \left[ A + (r-1)\frac{D}{2} \right] (p-q) \]

Collect A terms:

\[ A(q-r + r-p + p-q) = A(0) = 0 \]

Collect D/2 terms:

\[ \frac{D}{2} [ (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) ] \]

Evaluate bracket for D/2 terms:

\[ = pq - pr - q + r + qr - pq - r + p + pr - qr - p + q \]

All terms cancel each other out to yield 0.

\[ LHS = 0 + \frac{D}{2}[0] = 0 \]

Conclusion: Hence proved.

Question 2:

Determine the 10th term of a G.P. whose 3rd term is 24 and whose 6th term is 192.

Answer 2:

Let the first term be \( a \) and common ratio be \( r \).

We are given:

\[ a_3 = ar^2 = 24 \qquad \text{— (1)} \]

\[ a_6 = ar^5 = 192 \qquad \text{— (2)} \]

Divide eq (2) by eq (1):

\[ \frac{ar^5}{ar^2} = \frac{192}{24} \]

\[ r^3 = 8 \Rightarrow r = 2 \]

Substitute \( r=2 \) in eq (1):

\[ a(2^2) = 24 \Rightarrow 4a = 24 \Rightarrow a = 6 \]

Now we must find the 10th term (\( a_{10} \)):

\[ a_{10} = ar^9 = 6(2^9) = 6(512) = 3072 \]

Conclusion: The 10th term of the G.P. is 3072.