SEQUENCE
A succession of numbers f1, f2,…, fn formed according to some definite rule is called a sequence, f1, f2,…, fn are called 1st, 2nd,…, nth terms of the sequence. For example 1,3,5,7,… here each term of the sequence can be obtained by adding 2 to the preceding term. The Fibonacci sequence is given by a1 = 1, a2 = 1 and an+1 = an + an-1, n ≥ 2. The terms of this sequence are 1, 1, 2, 3, 5, 8, ...
Types of Sequence
There are two types of sequence:
- Finite sequence
- Infinite sequence
A sequence is said to be a finite or an infinite sequence according as it has finite or infinite number of terms.
Series
If a1, a2, a3,…, an,… is a sequence, then expression a1 + a2 + a3 + … + an + … is a series. In other words, a series is the sum of the terms of a sequence.
PROGRESSION
If the terms of the sequence follow a certain pattern, then the sequence is called progression. In particular there are three types of progressions:
- Arithmetic Progression
- Geometric Progression
- Harmonic Progression
ARITHMETIC PROGRESSION (A.P.)
A sequence is called an arithmetic progression if the difference of a term and the previous term is always same, i.e. an+1 – an = constant (= d) for all n ∈ ℕ.
The constant difference, generally denoted by d, is called the common difference.
nth Term and Sum of n Terms
If a is the first term and d the common difference, then the A.P. can be written as a, a + d, a + 2d, ...
The nth term an is given by an = a + (n – 1)d.
The sum Sn of the first n terms of such an A.P. is given by
Sn = n/2 [2a + (n – 1)d] = n/2 (a + l)
where l is the last term.
Arithmetic Mean(s)
If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then
b = (a + c) / 2 is the A.M. of a and c.
If a₁, a₂, ..., an are n numbers then the arithmetic mean (A) of these numbers is
A = 1/n (a₁ + a₂ + a₃ + ... + an)
GEOMETRIC PROGRESSION (G.P.)
A G.P. is a sequence whose first term is non-zero and each of whose succeeding term is r times the preceding term, where r is some fixed non-zero number, known as the common ratio of the G.P.
nth Term and Sum of n Terms
If a is the first term and r the common ratio, then G.P. can be written as a, ar, ar2, ... the nth term, an, is given by
an = arn−1.
The sum Sn of the first n terms of the G.P. is
Sn = a(rn − 1) / (r − 1), r ≠ 1
= na, r = 1
Geometric Means
If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P., then b = √ac is the geometric mean of a and c.
Arithmetico Geometric Progression
Suppose a₁, a₂, a₃,... is an A.P. and b₁, b₂, b₃,... is a G.P. Then the progression a₁b₁, a₂b₂, a₃b₃, ... is said to be an arithmetico–geometric progression (A.G.P). Hence, an arithmetico–geometric progression is of the form ab, (a+d)br, (a+2d)br², (a+3d)br³, ...
The nth term of the above A.G.P. is equal to the product of the nth term of A.P. (a, a + d, a + 2d, ...) and the G.P. (b, br, br², ..., brⁿ⁻¹, ...)
If –1 < r < 1, the sum of the infinite number of terms of the progression is
limn→∞ Sₙ = S = ab/1–r + dbr/(1–r)²
HARMONIC PROGRESSION (H.P.)
The sequence a1, a2, a3, …… an, … (ai ≠ 0) is said to be an H.P. if the sequence 1/a1, 1/a2, 1/a3, …… 1/an, …… is an A.P.
nth Term of H. P.
The nth term, an, of the H.P. is
an = 1 / [a + (n – 1)d] where a = 1/a1, and d = 1/a2 – 1/a1
Harmonic Means
If a and b are two non–zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P.
We have 1/H = ½ (1/a + 1/b) → H = 2ab / (a + b)
Some Important Results
- 1 + 2 + 3 + … + n = n(n + 1)
2 (sum of the first n natural numbers) - 12 + 22 + 32 + … + n2 = n(n + 1)(2n + 1)
6 (sum of squares of the first n natural numbers) - 13 + 23 + 33 + … + n3 = n2(n + 1)2
4 = (1 + 2 + 3 + … + n)2 (sum of cubes of first n natural numbers) - 1 + x + x2 + x3 + … = (1 − x)−1, if −1 < x < 1
- 1 + 2x + 3x2 + … = (1 − x)−2, if −1 < x < 1
INEQUALITIES
A.M. ≥ G.M. ≥ H.M.
Let a1, a2, ... , an be n positive real numbers, then we define their arithmetic mean (A), geometric mean (G), and harmonic mean (H) as:
- Arithmetic Mean (A):
A = (a1 + a2 + ... + an) / n - Geometric Mean (G):
G = (a1 a2 ... an)1/n - Harmonic Mean (H):
H = n / (1/a1 + 1/a2 + 1/a3 + ... + 1/an)
It can be shown that A ≥ G ≥ H. Moreover, equality holds at either place if and only if a1 = a2 = ... = an.
A, G, H form a G.P., i.e. G2 = AH.
Illustration 5:
If a, b, c are in H.P. and a > c > 0, then 1b-c - 1a-b
(A) is positive
(B) is zero
(C) is negative
(D) has no fixed sign.
Solution:
(A). b = H.M. of a and c < A.M. of a and c (as a and c are distinct)
⇒ b < (a + c)/2 ⇒ b - c < a - b ⇒ 1/(b-c) > 1/(a-b)
Formulas and Concepts
- If a is the first term and d the common difference of the A.P. Then nth term an is given by an = a + (n−1)d. The sum Sn of the first n terms of such an A.P. is given by Sn = n/2 [2a + (n−1)d] = n/2 (a + l) where l is the last term.
- If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then b = (a + c)/2 is the A.M. of a and c.
- If a is the first term and r the common ratio, then G.P. can be written as a, ar, ar2, ... the nth term, an, is given by an = arn−1. The sum Sn of the first n terms of the G.P. is
Sn = [a(rn − 1)] / (r − 1), r ≠ 1
Sn = na, r = 1 - If three terms are in G.P., then the middle term is called the geometric mean (G.M.) between the two. So if a, b, c are in G.P. then b = √(ac) is the geometric mean of a and c.
- The sequence a₁, a₂, a₃, ..., aₙ, ... (ai ≠ 0) is said to be an H.P. if the sequence 1/a1, 1/a2, 1/a3, ..., 1/an, ... is an A.P.
- The nth term, an, of the H.P. is an = 1 / [a + (n−1)d] where a = 1/a1, and d = 1/a2 − 1/a1.
- If a and b are two non–zero numbers, then the harmonic mean of a and b is a number H such that the numbers a, H, b are in H.P. We have 1/H = 1/2 (1/a + 1/b) ⇒ H = 2ab/(a + b).
- A.M. ≥ H.M. ≥ G.M.
- A, G, H form a G.P. i.e. G² = AH.
Solved Example
1. If first and (2n−1)th terms of an A.P., G.P., and H.P. are equal and their nth terms are a, b, c respectively, then
(A) a + c = 2b
(B) a + c = b
(C) a ≤ b ≤ c
(D) ac − b² = 0
Sol. (C). Let α be the first and β be the (2n−1)th terms of an A.P., G.P., and H.P., then α, a, β will be in A.P.; α, b, β will be in G.P.; α, c, β will be in H.P.
Hence a, b, c are respectively A.M., G.M., and H.M. of α and β.
Since A.M. ≥ G.M. ≥ H.M., a ≥ b ≥ c.
Again, \( a = \frac{α + β}{2} \), \( b^2 = αβ \), and \( c = \frac{2αβ}{α + β} \). Hence ac − b² = 0.
2. If a, b, c and d are distinct positive numbers in H.P., then
(A) a + b > c + d
(B) a + c > b + d
(C) a + d > b + c
(D) none of these
Sol. (C). Since b is the H.M. of a and c, \( \frac{a + c}{2} > b \) (A.M. > H.M.)
Again, c is the H.M. of b and d, \( \frac{b + d}{2} > c \) (A.M. > H.M.)
Adding, we get \( \frac{a + c}{2} + \frac{b + d}{2} > b + c \Rightarrow a + d > b + c \).
3. The number of terms common to the two A.P.'s 3, 7, 11, ... 407 and 2, 9, 16, ... ,709 is equal to
(A) 12
(B) 14
(C) 17
(D) none of these
Sol. (B)
Let number of terms in 2 A.P.'s be m and n respectively. Then,
407 = mth term of 1st A.P. and 709 = nth term of second A.P.
407 = 3 + (m – 1) × 4 and 709 = 2 + (n – 1) × 7
⇒ m = 102 and n = 102
So, each A.P. has 102 terms.
Let pth term of 1st A.P. be identical to qth term of 2nd A.P. Then,
3 + (p – 1) × 4 = 2 + (q – 1) × 7
4p – 1 = 7q – 5
⇒ 4p + 4 = 7q ⇒ 4(p + 1) = 7q
⇒ (p + 1)/7 = q/4 = k (say) ⇒ p = 7k – 1 and q = 4k
7k – 1 ≤ 102 and 4k ≤ 102
⇒ k ≤ 14(5/7) and k ≤ 25(1/2)
⇒ k ≤ 14
⇒ k = 1, 2, 3, ..., 14.
4. If a2 + b2, ab + bc and b2 + c2 are in G.P., then a, b, c are in
(A) G.P.
(B) A.P.
(C) H.P.
(D) none of these
Sol. (A)
a2 + b2, ab + bc and b2 + c2 are in G.P.; (ab + bc)2 = (a2 + b2)(b2 + c2)
a2b2 + b2c2 + 2ab2c = a2b2 + b4 + a2c2 + b2c2
b4 + a2c2 - 2ab2c = 0 ⇒ (b2 - ac)2 = 0 ⇒ b2 = ac ⇒ a, b, c are in G.P.
5. Sum of the series 12 – 22 + 32 – 42 + 52 – 62 + … + (n-1)2 – n2 is
(A) n(n-1)/2
(B) –n(n-1)
(C) (n2 – 1)/2
(D) (1 – n2)/2
Sol. (B)
12 – 22 + 32 – 42 + 52 – 62 + … + (n – 1)2 – n2
= (1 – 2)(1 + 2) + (3 – 4)(3 + 4) + … + (n – 1 – n)(n – 1 + n)
= –(1 + 2 + 3 + 4 + … + 2n – 1) = –(2n – 1)2n / 2 = –n(2n – 1)