Arithmetic Progression for SSC Maths 2014
Upcoming SlideShare
Loading in...5
×
 

Like this? Share it with your network

Share

Arithmetic Progression for SSC Maths 2014

on

  • 5,457 views

Arithmetic Progression for SSC Maths 2014, How to Prepare for SSC Maths, SSC Maths Preparation.

Arithmetic Progression for SSC Maths 2014, How to Prepare for SSC Maths, SSC Maths Preparation.

Statistics

Views

Total Views
5,457
Views on SlideShare
1,935
Embed Views
3,522

Actions

Likes
0
Downloads
25
Comments
0

10 Embeds 3,522

http://maharashtrassc.wordpress.com 2942
https://maharashtrassc.wordpress.com 520
http://www.ednexa.com 37
https://www.google.co.in 17
http://www.google.co.in 1
https://www.google.com 1
http://ednexassc.blogspot.in 1
http://ednexassc.blogspot.com 1
http://translate.googleusercontent.com 1
https://translate.googleusercontent.com 1
More...

Accessibility

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Arithmetic Progression for SSC Maths 2014 Document Transcript

  • 1. Arithmetic progression Sequence A sequence is a collection of numbers arranged in a definite order according to some definite rule. Each number in the sequence is called a term of the sequence. The number in the nth position is called the nth term and is denoted by ‘t n’. A sequence is usually denoted by t n  or t n and is read as sequence t n. Sum of the first n terms of a sequence If t n  is a sequence then we denote the sum of the first n terms of this sequence by Thus we see that Sn  t1  t 2  t 3  ....  t n S1  t1 S2  t1  t 2 S3  t1  t 2  t 3     Sn  t1  t 2  t 3    t n From the above we have
  • 2. S1  t1 S2  S1  t 2 S3  S2  t 3 Sn  Sn 1  t n  t n  Sn  Sn 1 Thus if S n is given we can find nth term and therefore any term of the sequence. If the number of terms in a sequence is finite then it is called a finite sequence, otherwise it is an infinite sequence. e.g. i) 1,3,5,7…. is an infinite sequence. ii) 4,8,12,….,400 is a finite sequence. Progressions A progression is a special type of sequence in which the relationship between any two consecutive terms is the same. There are three types of progressions namely Arithmetic Progression, Geometric Progression and Harmonic Progression.
  • 3. Arithmetic Progression (A.P.) An arithmetic progression is a sequence in which the difference between any two consecutive terms is always constant. The first term t1 is usually denoted by ‘a’ and the common difference is denoted by ‘d’. The value of d may be positive, negative or zero. For an A.P. t 2= t 1 + d = a + d t 3 = t 2 + d = a + d + d = a + 2d t 4 = t 3 + d = a + 2d + d = a +3d In general, t n – t n –1 = constant = d and so on…….
  • 4. General Term or nth Term of an A.P. Derivation of the formula for the general term: Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Here, …(1) t1=a t2–t1=d …(2) t3–t2=d …(3) :::::::::::::::::::::::::::::::::::::::::::::: t n –1 – t n – 2 t n – t n–1= d …(n –1) … (n) Adding the above equations, we get, t 1 + ( t 2 – t 1 ) + ( t 3 – t 2) + …+ ( t n – t n – 1) = a + d + d +….+ d (n – 1) times Therefore, t n = a + (n – 1) d. Thus, the general term, i.e. nth term of an A.P. with the first term ‘a’ and the common difference ‘d’ is given as t n = a + (n – 1) d.
  • 5. The sum of the first n terms of an A.P. Derivation of the formula Consider an A.P a, a + d, a + 2d,…, a (n – 1)d, where ‘a’ is the first term and ‘d’ is the common difference. The sum of the first n terms of an A.P. is Sn  a  a  d  ....  a  (n  2)d  a  (n  1)d ...(1) On reversing the terms and rewriting the expression, we get, Sn  a  (n  1)d  a  (n  2)d  ...  a  d   a ...(2) Adding equations (1) and (2), 2Sn  2a  (n  1)d  2a  (n  1)d  ....  2a  (n  1)d  (n times) 2Sn   2a  (n  1) d  Sn  n 2a  (n  1) d 2 The sum of the first n terms of an A.P. whose first term is ‘a’ and the common difference ‘d’ is Sn  Also, n  2a  (n  1) d  2
  • 6. n a  a  (n  1)d 2 n Sn   t1  t n ...where a  t1 and a  (n  1)d  t n 2 If we take t n  last term  , then n Sn  a   2 Sn  To know more please visit www.ednexa.com Or call 9011041155 for any help. - Team Ednexa