Here, you will learn what is arithmetic progression (ap) and formula for sum of ap series and properties of ap.
Let’s begin –
The indicated sum of the terms of a sequence. In the case of a finite sequence \(a_1\), \(a_2\), \(a_3\),………,\(a_n\) the corresponding series is \(a_1\) + \(a_2\) + \(a_3\) + ……… + \(a_n\) = \({\sum_{k=1}^{n}a_k}\). This series has a finite or limited numbers of terms and is called a finite series.
Formula for Sum of AP Series
A.P. is a sequence whose terms differ by a fixed number. This fixed number is called the common difference. If a is the first term & d the common difference, then A.P. can be written as
a, a + d, a + 2d, ………, a + (n – 1)d, ……..
(a) nth term of AP
\(T_n\) = a+(n-1)d,
where d = \(t_n\) – \(t_{n-1}\)
(b) The sum of the first n terms
\(S_n\) = \(n \over 2\)[2a + (n-1)d]
\(S_n\) = \(n \over 2\)[a + l]
where l is \(n^{th}\) term.
Also Read : Sum of GP Series Formula | Properties of GP
Note :
(i) \(n^{th}\) term of an A.P. is of the form An + B i.e. a linear expression in ‘n’, in such a case the coefficient of n is the common difference of the A.P. i.e. A.
(ii) Sum of first ‘n’ terms of an A.P. is of the form \(An^2\) + Bn i.e. a quadratic expression in ‘n’, in such case the common difference is twice the cofficient of \(n^2\). i.e. 2A
(iii) Also \(n^{th}\) term \(T_n\) = \(S_n\) – \(S_{n-1}\)
Example : If (x + 1), 3x and (4x + 2) are first three terms of an A.P. then its \(5^{th}\) term is-
Solution : (x + 1), 3x, (4x + 2) are in AP
=> 3x – (x + 1) = (4x + 2) – 3x => x = 3
=> a = 4, d = 9 – 4 = 5
=> \(T_5\) = 4 + (4)5 = 24
Hence, its \(5^{th}\) term is 24
Properties of AP
(a) If each term of an A.P. is increased, decreased, multiplied or divided by the some nonzero number, then the resulting sequence is also an A.P.
(b)
Three numbers in A.P. : a-d, a, a+d
Four numbers in A.P. : a-3d, a-d, a+d, a+3d
Five numbers in A.P. : a-2d, a-d, a, a+d, a+2d
Six numbers in A.P. : a-5d, a-3d, a-d, a+d, a+3d, a+5d etc.
(c) The common difference can be zero, positive or negative.
(d) \(k^{th}\) term from the last = (n – k+1)th term from the beginning (If total number of terms = n).
(e) The sum of the two terms of an AP equidistant from the beginning & end is constant and equal to the sum of first & last terms. => \(T_k\) + \(T_{n-k+1}\) = constant = a + l.
(f) Any term of an AP (except the first) is equal to half the sum of terms which are equidistant from it. \(a_n\) = (1/2)(\(a_{n-k}\) + \(a_{n+k}\)), k < n
(g) If a, b, c are in AP, then 2b = a + c.
Example : Four numbers are in A.P. If their sum is 20 and the sum of their squares is 120, then the middle terms are-
Solution : Let the numbers are a-3d, a-d, a+d, a+3d
given, a-3d + a-d + a+d + a+3d = 20
=> 4a = 20 => a=5
and \((a-3d)^2\) + \((a-d)^2\) + \((a+d)^2\) + \((a+3d)^2\) = 120
=> 4\(a^2\) + 20\(d^2\) = 120
=> 4 x \(5^2\) + 20\(d^2\) = 20 => \(d^2\) = 1 => d = \(\pm\)1
Hence numbers are 2,4,6,8 or 8,4,6,2
Related Questions
Prove that the sum of first n natural numbers is \(n(n+1)\over 2\)
