CBSE Class 11 Mathematics Sequences and Series Notes Set 01

Important Notes and Formulas

Definition

Sequence is a function whose domain is the set N of natural numbers.
Real Sequence: A sequence whose range is a subset of R is called a real sequence.
Series: If \( a_1, a_2, a_3, a_4, \ldots, a_n, \ldots \) is a sequence, then the expression \( a_1 + a_2 + a_3 + a_4 + a_5 + \ldots + a_n + \ldots \) is a series.
A series is finite or infinite according as the number of terms in the corresponding sequence is finite or infinite.
Progressions: It is not necessary that the terms of a sequence always follow a certain pattern or they are described by some explicit formula for the n^th term. Those sequences whose terms follow certain patterns are called progressions.

Geometric Progression (GP)

GP is a sequence of numbers whose first term is non-zero and each of the succeeding terms is equal to the preceding term multiplied by a constant. Thus in a GP the ratio of successive terms is constant. This constant factor is called the common ratio of the series and is obtained by dividing any term by the term which immediately precedes it. Therefore a, ar, ar², ar³, ar⁴, … is a GP with a as the first term and r as the common ratio.
(i) n^th term \( = ar^{n-1} \)
(ii) Sum of the first n terms, i.e. \( S_n = \frac{a(r^n - 1)}{r - 1} \), if r ≠ 1.
(iii) Sum of an infinite GP when |r| < 1: when n → ∞, \( r^n \to 0 \) if |r| < 1, therefore \( S_\infty = \frac{a}{1 - r} \) (|r| < 1).
(iv) If each term of a GP is multiplied or divided by the same non-zero quantity, the resulting sequence is also a GP.
(v) Any 3 consecutive terms of a GP can be taken as \( \frac{a}{r} \), a, ar; any 4 consecutive terms of a GP can be taken as \( \frac{a}{r^3}, \frac{a}{r}, ar, ar^3 \) and so on.
(vi) If a, b, c are in GP ⇒ b² = ac.

Properties of Geometric Progressions

1. If all the terms of a GP are multiplied or divided by the same non-zero constant, then it remains a GP with the same common ratio.
2. The reciprocals of the terms of a given GP form a GP.
3. If each term of a GP is raised to the same power, the resulting sequence also forms a GP.
4. In a finite GP the product of the terms equidistant from the beginning and the end is always the same and is equal to the product of the first and the last term.
5. Three non-zero numbers a, b, c are in GP if b² = ac.
6. If the terms of a given GP are chosen at regular intervals, then the new sequence so formed also forms a GP.

Arithmetic Mean

If three terms are in AP then the middle term is called the AM between the other two, so if a, b, c are in AP, b is the AM of a and c.
AM of any n positive numbers \( a_1, a_2, \ldots, a_n \) is \( A = \frac{a_1 + a_2 + a_3 + \ldots + a_n}{n} \)

n Arithmetic Means between Two Numbers

If a, b are any two given numbers and a, \( A_1, A_2, \ldots, A_n \), b are in AP, then \( A_1, A_2, \ldots, A_n \) are n AMs between a and b.
\( A_1 = a + \frac{b - a}{n + 1},\ A_2 = a + \frac{2(b - a)}{n + 1}, \ldots, A_n = a + \frac{n(b - a)}{n + 1} \)
i.e. \( A_1 = a + d,\ A_2 = a + 2d, \ldots, A_n = a + nd \), where \( d = \frac{b - a}{n + 1} \)
Note: The sum of n AMs inserted between a and b is equal to n times the single AM between a and b, i.e. \( \sum_{r=1}^{n} A_r = nA \), where A is the single AM between a and b.

Geometric Mean

If a, b, c are in GP, b is the GM between a and c: b² = ac, therefore \( b = \sqrt{ac} \); a > 0, c > 0.

n Geometric Means between a and b

If a, b are two given numbers and a, \( G_1, G_2, \ldots, G_n \), b are in GP, then \( G_1, G_2, G_3, \ldots, G_n \) are n GMs between a and b.
\( G_1 = a\left(\frac{b}{a}\right)^{\frac{1}{n+1}} = ar,\ G_2 = a\left(\frac{b}{a}\right)^{\frac{2}{n+1}} = ar^2, \ldots, G_n = a\left(\frac{b}{a}\right)^{\frac{n}{n+1}} = ar^n \), where \( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \)
Note: The product of n GMs between a and b is equal to the n^th power of the single GM between a and b, i.e. \( G_1 G_2 \ldots G_n = G^n \), where G is the single GM between a and b.

Relation between AM and GM of Two Numbers

Let A and G be the arithmetic and geometric means of two positive numbers a and b. Then
\( A = \frac{a + b}{2},\ G = \sqrt{ab} \)
1. A ≥ G (equality holds only when a = b), since \( A - G = \frac{(\sqrt{a} - \sqrt{b})^2}{2} \geq 0 \).
2. If A and G are respectively the arithmetic and geometric means between two positive quantities a and b, then the quadratic equation having a, b as its roots is x² − 2Ax + G² = 0.
3. If A and G are the AM and GM between two positive numbers, then the numbers are \( A \pm \sqrt{A^2 - G^2} \).

Solved Examples

Example 1. For the following G.P.'s find t_n
(i) 1, −4, 16, −64, …
(ii) \( \sqrt{3}, \frac{1}{\sqrt{3}}, \frac{1}{3\sqrt{3}}, \frac{1}{9\sqrt{3}}, \ldots \)
Answer:
(i) 1, −4, 16, −64, …
a = 1, \( r = \frac{-4}{1} = -4 \)
∴ \( t_n = ar^{n-1} = 1(-4)^{n-1} = (-4)^{n-1} \)
(ii) \( a = \sqrt{3} \), \( r = \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{1}{3} \)
∴ \( t_n = ar^{n-1} = \sqrt{3}\left(\frac{1}{3}\right)^{n-1} \)

Example 2. (i) Find three numbers in G.P. such that their sum is 35 and their product is 1000.
(ii) Find three numbers in G.P. such that their sum is \( \frac{13}{3} \) and the sum of their squares is \( \frac{91}{9} \).
Answer:
(i) Let the three numbers be \( \frac{a}{r} \), a, ar.
∴ \( \frac{a}{r} \times a \times ar = 1000 \) ⇒ a³ = 1000 ⇒ a = 10
\( a\left[\frac{1}{r} + 1 + r\right] = 35 \) ⇒ \( \frac{1}{r} + 1 + r = \frac{35}{10} \)
2(1 + r + r²) = 7r ⇒ 2r² − 5r + 2 = 0 ⇒ (r − 2)(2r − 1) = 0 ⇒ r = 2 or \( r = \frac{1}{2} \)
For r = 2: \( \frac{a}{r} = \frac{10}{2} = 5 \), ar = 10 × 2 = 20; the numbers are 5, 10, 20.
For \( r = \frac{1}{2} \): \( \frac{a}{r} = \frac{10}{1/2} = 20 \), \( ar = 10 \times \frac{1}{2} = 5 \); the numbers are 20, 10, 5.
(ii) Let \( \frac{a}{r} \), a, ar be three numbers in G.P.
\( \frac{a}{r} + a + ar = \frac{13}{3} \) … (i)    \( \frac{a^2}{r^2} + a^2 + a^2r^2 = \frac{91}{9} \) … (ii)
Squaring (i): \( \left(\frac{a^2}{r^2} + a^2 + a^2r^2\right) + 2a\left(\frac{a}{r} + a + ar\right) = \frac{169}{9} \)
\( \frac{91}{9} + 2a\left(\frac{13}{3}\right) = \frac{169}{9} \) ⇒ \( \frac{26a}{3} = \frac{78}{9} = \frac{26}{3} \) ⇒ a = 1
\( \frac{1}{r} + 1 + r = \frac{13}{3} \) ⇒ 3 + 3r + 3r² = 13r ⇒ 3r² − 10r + 3 = 0 ⇒ (r − 3)(3r − 1) = 0 ⇒ r = 3 or \( r = \frac{1}{3} \)
For r = 3: \( \frac{a}{r} = \frac{1}{3} \), ar = 3; the three numbers are \( \frac{1}{3} \), 1, 3.
For \( r = \frac{1}{3} \): \( \frac{a}{r} = 3 \), \( ar = \frac{1}{3} \); the three numbers are 3, 1, \( \frac{1}{3} \).

Example 3. If x, y and z are p^th, q^th and r^th terms of a G.P. respectively, then show that \( x^{q-r} \cdot y^{r-p} \cdot z^{p-q} = 1 \)
Answer:
Let A be the first term and R be the common ratio of the given G.P. Then
x = p^th term ⇒ \( x = AR^{p-1} \); y = q^th term ⇒ \( y = AR^{q-1} \); z = r^th term ⇒ \( z = AR^{r-1} \)
L.H.S. \( = \left(AR^{p-1}\right)^{q-r} \cdot \left(AR^{q-1}\right)^{r-p} \cdot \left(AR^{r-1}\right)^{p-q} \)
\( = A^{(q-r)+(r-p)+(p-q)} \cdot R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)} \)
\( = A^0 \cdot R^{pq - pr - q + r + qr - pq - r + p + pr - qr - p + q} = A^0 R^0 = 1 \) = R.H.S.

Example 4. If for a sequence \( t_n = \frac{2^{n-2}}{5^{n-3}} \), show that the sequence is a G.P. Find its first term and the common ratio.
Answer:
\( t_n = \frac{2^{n-2}}{5^{n-3}} = \frac{2^n \cdot 5^{3}}{2^{2} \cdot 5^n} = \left(\frac{5^3}{2^2}\right)\left(\frac{2}{5}\right)^n = \frac{125}{4}\left(\frac{2}{5}\right)^n \)
\( t_{n+1} = \frac{125}{4}\left(\frac{2}{5}\right)^{n+1} \)
\( \frac{t_{n+1}}{t_n} = \frac{\frac{125}{4}\left(\frac{2}{5}\right)^{n+1}}{\frac{125}{4}\left(\frac{2}{5}\right)^n} = \frac{2}{5} \) = constant
Hence the sequence is a GP.
\( t_1 = a = \frac{125}{4} \times \frac{2}{5} = \frac{25}{2} \), and common ratio \( r = \frac{t_{n+1}}{t_n} = \frac{2}{5} \)

Example 5. For a G.P.
(i) If a = 2, r = 3, S_n = 242, find n.
(ii) If S₃ = 125, S₆ = 152, find r.
Answer:
(i) a = 2, r = 3, S_n = 242
\( S_n = a\left[\frac{r^n - 1}{r - 1}\right] \) ⇒ \( 242 = 2\left[\frac{3^n - 1}{3 - 1}\right] \)
242 = 3ⁿ − 1 ⇒ 243 = 3ⁿ ⇒ 3⁵ = 3ⁿ ⇒ n = 5
(ii) \( S_3 = a\left[\frac{r^3 - 1}{r - 1}\right] \) and \( S_6 = a\left[\frac{r^6 - 1}{r - 1}\right] \)
\( \frac{S_6}{S_3} = \frac{r^6 - 1}{r^3 - 1} \) ⇒ \( \frac{152}{125} = \frac{r^6 - 1}{r^3 - 1} \)
By dividendo: \( \frac{152 - 125}{125} = \frac{r^6 - 1 - (r^3 - 1)}{r^3 - 1} = \frac{r^3(r^3 - 1)}{r^3 - 1} \)
\( \frac{27}{125} = r^3 \) ⇒ \( \left(\frac{3}{5}\right)^3 = r^3 \) ∴ \( r = \frac{3}{5} \)

Example 6. Find the sum to n terms.
(i) 0.9 + 0.99 + 0.999 + …
(ii) 0.5 + 0.55 + 0.555 + …
Answer:
(i) S_n = [0.9 + 0.99 + 0.999 + …] = [(1 − 0.1) + (1 − 0.01) + (1 − 0.001) + …]
= [(1 + 1 + 1 + …) − (0.1 + 0.01 + 0.001 + …)]
= n − (sum of n terms of a GP with a = 0.1 and r = 0.1)
\( S_n = n - \left[\frac{(0.1)(1 - 0.1^n)}{1 - 0.1}\right] = n - \left[\frac{0.1}{0.9}\left(1 - \frac{1}{10^n}\right)\right] = n - \frac{1}{9}\left[1 - \frac{1}{10^n}\right] \)
(ii) S_n = 0.5 + 0.55 + 0.555 + … = 5(0.1 + 0.11 + 0.111 + …)
\( = \frac{5}{9}(0.9 + 0.99 + 0.999 + \ldots) = \frac{5}{9}[(1 - 0.1) + (1 - 0.01) + (1 - 0.001) + \ldots] \)
\( = \frac{5}{9}[(1 + 1 + 1 + \ldots) - (0.1 + 0.01 + 0.001 + \ldots)] \)
\( = \frac{5}{9}\left\{n - \frac{(0.1)(1 - 0.1^n)}{1 - 0.1}\right\} = \frac{5}{9}\left\{n - \frac{1}{9}\left[1 - \frac{1}{10^n}\right]\right\} \)

Example 7. Determine whether the sum to infinity of the following G.P.s exists; in case it exists, find it.
(i) 1, 2, 4, 8, 16, …
(ii) \( 1, \frac{3}{2}, \frac{9}{4}, \frac{27}{8}, \ldots \)
(iii) \( 5, \frac{-5}{2}, \frac{5}{4}, \frac{-5}{8}, \frac{5}{16}, \ldots \)
Answer:
(i) a = 1, r = 2: \( S_n = a\left(\frac{1 - r^n}{1 - r}\right) = \frac{1 - 2^n}{1 - 2} = 2^n - 1 \)
r = 2 > 1 ∴ The sum to infinity does not exist.
(ii) a = 1, \( r = \frac{3}{2} \): \( S_n = \frac{1 - \left(\frac{3}{2}\right)^n}{1 - \frac{3}{2}} = -2\left[1 - \left(\frac{3}{2}\right)^n\right] = 2\left(\frac{3}{2}\right)^n - 2 \)
\( r = \frac{3}{2} > 1 \) ∴ The sum to infinity does not exist.
(iii) a = 5, \( r = \frac{-5/2}{5} = -\frac{1}{2} \)
\( S_n = 5\left[\frac{1 - \left(-\frac{1}{2}\right)^n}{1 - \left(-\frac{1}{2}\right)}\right] = 5\left[\frac{1 - \left(-\frac{1}{2}\right)^n}{\frac{3}{2}}\right] = \frac{10}{3}\left[1 - \left(-\frac{1}{2}\right)^n\right] \)
\( |r| = \left|-\frac{1}{2}\right| < 1 \), so \( \left(-\frac{1}{2}\right)^n \to 0 \) as n → ∞
∴ \( S_\infty = \frac{10}{3}[1 - 0] = \frac{10}{3} \)

Example 8. For a sequence, if S_n = 7(4ⁿ − 1), find t_n and show that the sequence is a G.P.
Answer:
S_n = 7(4ⁿ − 1)
t_n = S_n − S_n−1 = 7[4ⁿ − 1] − 7[4ⁿ⁻¹ − 1]
\( = 7(4^n) - 7 - 7\frac{(4)^n}{4} + 7 = 7 \times 4^n\left[1 - \frac{1}{4}\right] = 7 \times 4^n \times \frac{3}{4} = 21[4^{n-1}] \)
\( r = \frac{t_{n+1}}{t_n} = \frac{21[4^{n}]}{21[4^{n-1}]} = 4 \)
The common ratio is constant. Hence the given sequence is a GP.

Example 9. Find the sum of 1 + (1 + x) + (1 + x + x²) + … + (1 + x + x² + … + xⁿ⁻¹)
Answer:
S_n = 1 + (1 + x) + (1 + x + x²) + … + (1 + x + x² + … + xⁿ⁻¹)
T_r = 1 + x + x² + … + x^r−1. This is a GP with first term 1 and common ratio x.
∴ \( T_r = \frac{1(1 - x^r)}{1 - x} \)
\( S_n = \sum_{r=1}^{n} \frac{1 - x^r}{1 - x} = \frac{1}{1 - x}\sum_{r=1}^{n} 1 - \frac{1}{1 - x}\sum_{r=1}^{n} x^r \)
\( = \frac{1}{1 - x}(n) - \frac{1}{1 - x}\left[\frac{x(1 - x^n)}{1 - x}\right] = \frac{n}{1 - x} - \frac{x(1 - x^n)}{(1 - x)^2} \)