Sequences and Series

Arithmetic sequence/progression

The sum of the first n terms of an arithmetic sequence
The sum of the first n natural numbers
Arithmetic sequences, examples
Sequences
Sequence or progression is an ordered set of numbers, either finite or infinite, such that each term in a sequence can be indexed meaning, can be written as an algebraic function of its position in the sequence.
A finite sequence has a definite number of terms while an infinite sequence has an infinite number of terms.
Arithmetic sequence/progression
An arithmetic progression is a sequence of numbers in which the difference between any two successive numbers is constant.
The difference d between successive terms is called common difference.
Therefore, an arithmetic sequence can be written as
a1a2a3a4, . . . , an -1an, . . .    or    a1a1 + da1 + 2da1 + 3d, . . . , an -1,  an, . . .
where,    a2 = a1 + d,
a3 = a1 + 2d,
a4 = a1 + 3d, and so on,
thus, the formula for the nth term or the general term of an arithmetic sequence is
 an = a1 + (n - 1) · d.
The sum of the first n terms of an arithmetic sequence
When deriving the formula for the sum of the first n terms of an arithmetic sequence we use the fact that the sum of every pair of symmetric terms of the sequence is the same, as is the sum of the first and last terms in the sequence the same as, the sum of the second and second to last terms, and so on.
Therefore, by adding the sum of all terms of a sequence to the sum of the same terms written in the inverse
order, that is            Sn = a1 + a2 + a3 + . . . + an -2 + an -1 + an,
and      Sn = an + an -1 + an -2 + . . . + a3 + a2 + a1
then,  2Sn = (a1 + an) + (a2 + an -1) + (a3 + an -2) + . . . + (a3 + an -2) + (a2 + an -1) + (a1 + an)
since the right side of the equation represents the sum of n partial sums each of the same value,
a1 + an   or   2a1 + (n - 1) · d
then, the sum of the first n terms of the arithmetic sequence
 Sn = (n/2) · (a1 + an)    or    Sn =  (n/2) · [2a1 + (n - 1) · d]
Let find the sum of the arithmetic sequence        3,   6,   9,  12,  15
we add the same sequence in inverse order      15,  12,  9,    6,   3
it follows that    2Sn = (3 + 15) + (6 + 12) + (9 + 9) + (12 + 6) + (15 + 3) = 5 · 18 = 90,
thus,     2Sn = n · (a1 + an) = 5 · (3 + 15)   oSn = (n/2) · (a1 + an) = 5/2 · 18 = 45.
The sum of the first n natural numbers
As natural numbers represents the arithmetic sequence with the first term a1 = 1 and common difference
d = 1, we substitute these values into the formula for the sum of the first n terms of the arithmetic sequence,
a1 = 1 and  d = 1   =>   Sn =  (n/2) · [2a1 + (n - 1) · d] = (n/2) · [2 · 1 + (n - 1) · 1],
 to obtain, Sn =  n · (n + 1)/2 the formula for the sum of the first n natural numbers.
Arithmetic sequences, examples
Example:  The sum of three successive terms of an arithmetic sequence is 33 and the product is 1232.
Find the greatest term.
Solution:  Since,  (1)  a1 + a2 + a3 = 33,             and                              (2)  a1 · a2 · a3 = 1232
then   a1 + a1 + d + a1 + 2d = 33                                   a1 · (a1 + d) · (a1 + 2d) = 1232
3a1 + 3d = 33,
a1 = 11 - d    =>    (2)  (11 - d) · (11 - d + d ) · (11 - d + 2d ) = 1232
d =   =>           a1 = 11 - d = 11 - 3,                                                121 - d 2 = 112,
a1 = 8.                                                                                d 2 = 9,    d = 3,
Thus, the given sequence is   8,  11,  14.
Example:  If the sum of the first fifteen terms of an arithmetic sequence is 0 and the first term is 21, find the tenth term of the sequence.
Solution:    S15 = 0 and a1 = 21,      Sn = (n/2) · (a1 + an)                              an = a1 + (n - 1) · d
a10 = ?                          S15 = (15/2) · (21 + a15) = 0                 - 21 = 21 + 14 · d
a15 = - 21,           14 d = - 42,    d = - 3
a10 a1 + 9d
a10 21 + 9 · ( - 3) = - 6,    a10 = - 6.
Example:  Find the sum of all natural numbers greater than 30 and smaller than 330 whose last digit is 1.
Solution:    Given numbers form the sequence   31, 41, 51, . . . , 291, 301, 311, 321.
Therefore,   a1 = 31,   an = 321  and  d = 10
as       an = a1 + (n - 1) · d
then     321 = 31 + (n - 1) · 10                                 Sn = (n/2) · (a1 + an)
290 = (n - 1) · 10                                         S30 = 30/2 · (31 + 321)
n - 1 = 29,     n = 30.                                     S30 = 15 · 352 = 5280.
Example:  Find the first term and number of terms of an arithmetic sequence with an = 15Sn = 64 and
common difference d = 2.
Solution:  Use the formulas for an and Sn to form a system of two equations in two unknowns,
(1)  an = a1 + (n - 1) · d
(2)  Sn =  (n/2) · [2a1 + (n - 1) · d]

(1)  a1 + (n - 1) · 2 = 15,                 =>           a1 = 17 - 2n
(2)   (n/2) · [2a1 + (n - 1) · 2] = 64,    =>            n · (a1+ n - 1) = 64
(1)   a1 = 17 - 2n          =>     (2)     n2 + a1n - n = 64
n2 + (17 - 2n) · n - n = 64,
n2 - 16n + 64 = 0,
n1,2 = 8 ± Ö 64 - 64,    n = 8,   a1 = 17 - 2 · 8 = 1.
Therefore, the sequence is  1, 3, 5, 7, 9, 11, 13, 15, . . . , where  a8 = 15 and  S8 = 64.
Example:  Which arithmetic sequence has the property that the sum of its first five terms is 35 and the sum of the second and the sixth term is 20.
Solution:  Use the formulas for Sn and  ato form a system of two equations,
Sn =  (n/2) · [2a1 + (n - 1) · d]                    =>    (1)   (5/2) · [2a1 + 4d] = 35,                a1 + 2d = 7
an = a1 + (n - 1) · d,         a2 + a6 = 20    =>    (2)    (a1 + d) + (a1 + 5d) = 20,      2a1 + 6d = 20
(2)   a1 + 3d = 10
(2) - (1)    (a1 + 3d) - (a1 + 2d) = 10 - 7,      d = 3
(2)   a1 + 3d = 10   =>   a1 = 10 - 3d = 10 - 3 · 3 = 1,    a1 = 1.
Thus, the sequence is  1, 4, 7, 10, 13, 16, 19, . . .
Example:  How many natural numbers, divisible by 7, lie between 0 and 100, and what is their sum.
Solution:  The sequence of numbers divisible by 7 is;   7, 14, 21, 28, . . . , 84, 91, 98
and since   an = a1 + (n - 1) · d,    then     98 = 7 + (n - 1) · 7 | ¸ 7
Sn = n/2 · (a1 + an),                    n = 14,
S14 = 14/2 · (7 + 98) = 7 · 105 = 735.
Example:  Find n and Sn of an arithmetic sequence if given are;  a1, an, and d.
Solution:  Since,   an = a1 + (n - 1) · d    then    an - a1 = (n - 1) · dor   n - 1 = (an - a1) / d
n = (an - a1 + d) / d
and    Sn = (n/2) · (a1 + an)
by substituting    n = (an - a1 + d) / d    obtained is  Sn = (an - a1 + d) · (a1 + an) / (2d)
Example:  Find d and n of an arithmetic sequence if given are;  a1, an, and Sn.
Solution:  Since,   an = a1 + (n - 1) · d           and          Sn = (n/2) · (a1 + an)
d = (an - a1) / (n - 1)        <=            n = (2Sn) / (a1 + an)
d = (an - a1) / [(2Sn) / (a1 + an) - 1] = (an2 - a12) / (2Sn - a1 - an)
Example:  Derive the formula for the nth odd natural number and the sum of the first n odd natural numbers.
Solution:  The sequence of odd natural numbers is   1, 3, 5, 7, . . . , an, . . .
using       an = a1 + (n - 1) · d
an = 1 + (n - 1) · 2         and since         Sn = (n/2) · (a1 + an)
an = 2n - 1       =>   Sn           then         Sn = (n/2) · [1 + (2n - 1)]   so  Sn = n2.
Example:  In a sequence of natural numbers divisible by 7 find the one that equals the one seventeenth of the sum of all natural numbers that precede this one.
Solution:  A natural number divisible by 7 is 7n, where n Î N, so the sequence of the numbers is
7, 14, 21, 28, 35, 42, 49,  . . .
and since the formula for the sum of the first n natural numbers    Sn n · (n + 1)/2,
and the sum of natural numbers that precede the number divisible by 7 is,    (7n + 1)[(7n - 1) + 1] / 2
then, the following equation meets the given condition
7n = (1/17) · 7n(7n - 1) / 2    thus,   7n = 35.
Intermediate algebra contents