Resources | Subject Notes | Mathematics
An arithmetic progression is a sequence of numbers where the difference between any two consecutive terms is constant. This constant difference is called the common difference, denoted by 'd'.
The formula for the nth term (an) of an arithmetic progression is:
$a_n = a_1 + (n - 1)d$
where:
The sum (Sn) of the first n terms of an arithmetic progression is given by:
$S_n = \frac{n}{2}(a_1 + a_n)$
Alternatively, it can be expressed as:
$S_n = \frac{n}{2}(2a_1 + (n - 1)d)$
Find the sum of the first 20 terms of the arithmetic progression 3, 8, 13, 18,...
Here, $a_1 = 3$, $d = 8 - 3 = 5$, and $n = 20$
$S_{20} = \frac{20}{2}(2(3) + (20 - 1)5)$
$S_{20} = 10(6 + 19 \times 5)$
$S_{20} = 10(6 + 95)$
$S_{20} = 10(101)$
$S_{20} = 1010$
A geometric progression is a sequence of numbers where each term is multiplied by a constant value to obtain the next term. This constant value is called the common ratio, denoted by 'r'.
The formula for the nth term (an) of a geometric progression is:
$a_n = a_1 \times r^{n - 1}$
where:
The sum (Sn) of the first n terms of a geometric progression is given by:
$S_n = a_1 \times \frac{1 - r^n}{1 - r}$
This formula is valid when $r \neq 1$.
If $r = 1$, then $S_n = n \times a_1$
Find the sum of the first 10 terms of the geometric progression 2, 6, 18, 54,...
Here, $a_1 = 2$, $r = \frac{6}{2} = 3$, and $n = 10$
$S_{10} = 2 \times \frac{1 - 3^{10}}{1 - 3}$
$S_{10} = 2 \times \frac{1 - 59049}{-2}$
$S_{10} = 2 \times \frac{-59048}{-2}$
$S_{10} = 2 \times 29524$
$S_{10} = 59048$
The binomial expansion provides a formula for expanding expressions of the form $(a + b)^n$, where n is a non-negative integer.
The general term in the binomial expansion of $(a + b)^n$ is given by:
$T_{k+1} = \binom{n}{k} a^{n-k} b^k$
where:
Expand $(x + y)^3$ using the binomial expansion.
Here, $a = x$, $b = y$, and $n = 3$
$T_{k+1} = \binom{3}{k} x^{3-k} y^k$
Therefore, $(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$