1 Jan 2016

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Arithmetic Progression Made Easy

An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant.

For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1.

Second example: the sequence 3, 5, 7, 9, 11,... is an arithmetic progression
with common difference 2.
Third example: the sequence 20, 10, 0, -10, -20, -30, ... is an arithmetic progression
with common difference -10.

 

Notation


We denote by d the common difference.

By an we denote the n-th term of an arithmetic progression.

By Sn we denote the sum of the first n elements of an arithmetic series.
Arithmetic series means the sum of the elements of an arithmetic progression.

Properties


a1 + an = a2 + an-1 = ... = ak + an-k+1

and
an = ½(an-1 + an+1)

Sample: let 1, 11, 21, 31, 41, 51... be an arithmetic progression.

51 + 1 = 41 + 11 = 31 + 21
and
11 = (21 + 1)/2
21 = (31 + 11)/2...




If the initial term of an arithmetic progression is a1 and the common difference of successive members is d, then the n-th term of the sequence is given by
an = a1 + (n - 1)d, n = 1, 2, ...





The sum S of the first n numbers of an arithmetic progression is given by the formula:
S = ½(a1 + an)n

where a1 is the first term and an the last one.

or
S = ½(2a1 + d(n-1))n





























 

Arithmetic Progression Problems


1) Is the row 1,11,21,31... an arithmetic progression?
Solution: Yes, it is an arithmetic progression. Its first term is 1 and the common differnece is 10.




2) Find the sum of the first 10 numbers of this arithmetic series: 1, 11, 21, 31...
Solution: we can use this formula S = 1/2(2a1 + d(n-1))n
S = 1/2(2.1 + 10(10-1))10 = 5(2 + 90) = 5.92 = 460




3) Try to prove that if the numbers 1/(c + b), 1/(c + a), 1/(a + b) form an arithmetic progression then the numbers a2, b2, c2 form an arithmetic progression too.

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