Sequence and series are fundamental topics in arithmetic and quantitative aptitude. A sequence is an ordered list of numbers (or objects) that follow a specific rule. A series is the sum of the terms of a sequence. Common examples include arithmetic progressions and geometric progressions, which are widely used in problem solving for competitive examinations.
Let the terms of a sequence be denoted by a1, a2, a3, a4, ... where the index 1, 2, 3, 4, ... denotes the position of each term.
A sequence may be finite or infinite. If a1, a2, ..., aN is a finite sequence, the corresponding finite series is
SN = a1 + a2 + a3 + ... + aN
If the sequence is infinite, the associated series may be finite (convergent) or infinite (divergent) depending on whether the infinite sum converges.
In CAT-oriented problems, infinite series-when discussed-are typically limited to geometric progressions with |r| < 1.
An arithmetic progression is a sequence in which each term after the first is obtained by adding a fixed number, called the common difference d, to the previous term. If a is the first term, the sequence is
a, a + d, a + 2d, a + 3d, ...
The nth term (Tn) of an AP is
Tn = a + (n - 1)d
The last term l (if there are n terms) satisfies
l = a + (n - 1)d
The sum of the first n terms (Sn) of an AP is
Sn = n/2 × [2a + (n - 1)d] = n/2 × (a + l)
A geometric progression is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed number, called the common ratio r. If a is the first term, the sequence is
a, ar, ar2, ar3, ...
The nth term (Tn) of a GP is
Tn = a rn-1
The sum of the first n terms (Sn) for r ≠ 1 is
Sn = a × (1 - rn) / (1 - r)
If |r| < 1, the infinite geometric series converges and its sum is
S∞ = a / (1 - r)
Note: In CAT, infinite series questions are limited only to geometric progressions with |r| < 1.
A sequence is in harmonic progression if the reciprocals of its terms form an arithmetic progression. For example, if bn = 1/an is an AP, then an is an HP.
If a1, a2, a3,...... are in HP, then 1/a1, 1/a2, 1/a3, ...... are in AP.
Note: Problems involving HP are solved by converting the HP into an AP by taking reciprocals of the terms.
The Fibonacci sequence is defined by the recurrence
F0 = 0, F1 = 1, and Fn = Fn-1 + Fn-2 for n ≥ 2.
The sequence begins 0, 1, 1, 2, 3, 5, 8, 13, ...
Note: For CAT, Fibonacci problems are solved using the recurrence relation or pattern recognition. Closed-form formulas are not required.

*Here, a = first term, d = common difference, r = common ratio, n = position of term, l = last term.
A sequence lists elements in order; a series sums elements of a sequence. A sequence may be finite or infinite. A series formed from an infinite sequence may converge (sum has a finite value) or diverge.

Question 1: If 4,7,10,13,16,19,22......is a sequence, Find:
(a) Common difference
(b) nth term
(c) 21st term
Sol:
(a) The common difference = 7 - 4 = 3
(b) The nth term of the arithmetic sequence is denoted by the term Tn and is given by Tn = a + (n-1)d, where "a" is the first term and d is the common difference.
Tn = 4 + (n - 1)3 = 4 + 3n - 3 = 3n + 1
(c) 21st term as: T21 = 4 + (21-1)3 = 4+60 = 64.
Q2: Consider the sequence 1, 4, 16, 64, 256, 1024..... Find the common ratio and 9th term.
Sol: The common ratio (r) = 4/1 = 4
The preceding term is multiplied by 4 to obtain the next term.
The nth term of the geometric sequence is denoted by the term Tn and is given by Tn = ar(n-1) where a is the first term and r is the common ratio.
Here a = 1, r = 4 and n = 9
So, 9th term is can be calculated as T9 = 1* (4)(9-1)= 48 = 65536.
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