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# arthimatic progressions

Published on: Mar 3, 2016
Published in: Education
Source: www.slideshare.net

#### Transcripts - arthimatic progressions

• 1. .‘ I L’. J" ' r’_‘u ‘ V. ‘ , ". >.. . 1 5’ .7‘ " . _ ‘C 14.’ , , I "I V ’ 9}’ ’ to I . -. - _ -6’; “nary . _.. ’.t- T l . 1| ’| ._
• 2. Cl Arithmetic Sequence is a sequence of numbers such that the difference between the consecutive terms is constant. 15 is an arithmetic progression with common difference of 2. Fr 2,6,18,S4(next term to the term is to be obtained by multiplying by 3. Arithmetic Sequence
• 3. c A finite portion of an brithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is calledlan Arithmetic series. ~ The behavior of the afithmetic progression depends on the common difference gl. If the common difference is: ~/ Positive, the membeds (terms) will grow towards positive jnfifnity. ’*“~/ 'N‘eg§tive", * ‘t‘h"e em l5eFs’(terms) will‘g row fnuua rrle ncanaﬁIn infi niht
• 4. C] lfwr: take first term ofsm AP asz and Common Difference as 4’. Then» nth term of that Al’ will be Ari: til»! LL > For instance-~ 3, 7, 11, 1'3, 19 J =4 :2 =3 Notice in this sequence that if we find the tiiffttrcntte between any term and the term before 1t we always get -1. ' 4 is then called the cnmmandiffezcncc and is denoted with the letter d. " 1'0 [it to the next term in the sequence we would add ’i so a recursive fonnula for 18 sequence is: CI The first term in the sequence would be 4:, which is sometimes just written as 11.
• 5. Illustrative example for A. P. = d, where d= l
• 6. Ari: tin1;. e'ti. c Progression / If the initial term of an arithmetic progression is Q; and the common difference of successive members is g, then the gig term of the sequence (a, ,) is given by: V An: a1+(n—1)d *\$+, ’<Inc{in general 41"‘ ("'m}’[ i)
• 7. a _ _ “%_ : — A Ar. To ﬁnd the nth Term ofan . P. ( _’; ‘s ‘ M ‘ U“ Lei: us consider an v étlw 'i‘irst ‘ten and common di'ffet'ence 'd‘ men 3'4-. —. W * ‘»‘c: : '3 . 7 " . : ft ﬂow 3;; :fAf; fv""'“w"\$4; ’ . .f. .3« ‘4:-. ———‘~" 7 ~25 ‘- 1 ; « . ' . . v . ‘ *~ ~ —. .~: . ' ; «:>; =9, —> a «_ _ .7.‘ 5n. ‘.r,3;e‘£, =r’qf4 _ ~' 41;, 3, v I ‘-"}= :,, .
• 8. Let a=2, d=2, n=12,find An , An= a+(n-1)d “L =2+(12-1)2 A =2+(11)2 »/ =2+22 , ' Therefore, An=24 Hence solved. I~ r ’ ' r r r‘ f r‘ r r r r r r r -- * " . / _/ ,/ _,/ 1/ , / _/ ,/ _/ __,2 -/ J , / J. _/ _/ ,,2 _/ ,,/
• 9. To chec| < that a given term is in A. P. or not.
• 10. Pr©bD@m 3.. Find 10"‘ term of A. P. 12, 18, 24, 30.. f :1 3 i i. . 5.. . iv P" P, x. -« K ’; ;‘: ;i "u. -.1‘. :n is :1: 1:2. «"1 . , ' . _ , ' . I __ 1‘) . » . _ c; zr‘. .mr. >u d,3;. ;e1L:1L, c 1:. » (2 — 1«_~: - A‘ i I -; i. :2 = .«'; '-%''. fl. i’_)-J_)i'') ii . ..1 ii [I7 1!
• 11. j: _ *““ ‘, ‘‘’'r ’'3‘‘‘ ’‘ ~". t/V“ ‘ _V_, - . . . .._. ... ‘,. ‘ -_. ‘». .« . .._. - _ ‘. -__ '_J’. ..____. To find the Arithmetic Mean Between Two Numbers. Let A be the AM between a and b. A, A, b are in AP (A-a)= (b-A) A=1/2(a+b) AM between a and b A=1/2(a+b) 3.lj1F- _’}f4:': ,-3*. ’- j_3jl. -' ‘J? ?-
• 12. Lo’ Some convenient methods to determine the AP ” w v “I1? i'| ‘—" i’ ii sei, ’-_| i_'€. .=-5 ‘Hill-: :.i ~. .-W i. 3 numbers in an AP as (a-ti), a, (a-2-ti), ii. 4 numbers in an AP as (a—3d), (a-o’), a, (a+d), (a+3d); iii. 5 numbers in an AP as (a—2d), (a—d), a, (a+d), (a+2d)
• 13. Q---- The sum of three numbers in an AP is 21 8.: their product is 231. Find the numbers. Let the required Numbers be (a-d), a, (a+d) Then (a-d)+ a+ (a+d)=21 / A:7 / And (a-d) a (a+d)=231 / A(a1-dz) =231 ~/ Substituting the value of ‘a’ V 7(49-d2)=231 ~/ 7d2= (343-231): 112 ~/ D2: 16 V 0: L3.
• 14. A i Sum of n-term of an ap l The sum of n terms, we find as, Sum = n X [(first term + last tei'm)/ 2] Now last is: in will be = a (n-1) cl Therefore, ~/ Sn= ‘/2n[2a+(n-1)d] It can also be written as V S, ,=‘/2n[a+a, ,]
• 15. Problem 1. Find the sum of 30 Terms of given A. P. 12 + 20 + 28 + . Solu‘iion (3-i~. /en AP i512 20 28 335) _ Its first term is o = 12 ' 1 ' ‘_- 1 Common difference is d : 20 - 12 = 8 - -0‘ T an ’ 5/; x_3O [ 2x}2 + (36-1)}: s] ' 1 = 15 [.24 + 29 X8] ' = 1s[24 + 232] = 15 x 246 = 3690 ‘ ‘ The. su_rﬁ to n fe'rr_n_s. -sf an arithmeticprfoigréssion. ' ‘ _' ; ,sn§'/2 n‘[L2'a-i: (n-.11). ?-]. .~"_ - ‘_
• 16. Practice Q1: Find the sum of the following APs: (i) 2, 7, 12, . ... .. ,to 10 terms 4 2‘ ‘:1 '2 I1} 7 i A U5 1 1 l I , Q3: Given a = 2, d = 8, 5,, = 90, find n and an. :1 '- . - ij I I 7 V” ~-*, ... — .1 memes. . 1 I . ll *. r F ‘+1.. 1;! ..'_**v e. "' . @433 5&1 l mu’? - ab. -‘i l
• 17. frrrreeiiczuuam iﬁ1l1ljiifﬁ)i§i@U° @711‘ :1u: :,, M3,, Solution. 9) n_ 1 = 80 1) First term is a = 100 , anzi-00 10) n=80+1 2) Common diﬂ’erence is d 11) n = 81 = 105 -100 = 5 . _ ~/ Hence the no. of terms are - 3) nth term 15 an = 21 + (n- 1)(] -81. 4) 500 = 100 + (n-1)5 5) 50o — 100 = 5(n — 1) 6) 400 = 5(n - 1) 7) 5(n — 1) = 400 8) n — 1 = 400/5
• 18. Question: 1. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. Solution: an :14? = a+7 31 = ‘.-149-147: 2‘ =0 -'1 '3 rJl: |‘-/ ‘I. 1:22 — — Question: 2. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. I { ‘ rag " 3: Solution: Here 0 2 4 ‘arm; 3 2 H - -'- : 10 So, .: _e, =1C‘+-\$. .5i'J: :.1'. :l it- 5. i Hnn. ‘ \$7,), :+'l"= +'il| "i: ‘f-l }l| ‘=: ‘f"-EH I
• 19. L- » . . ‘' ya‘ ‘ 2 6' I ll “ v . ,;_s ’. -_. ’;-: -t *. , _ I ‘1 r , .-~ ' .0 ' . .1‘ ‘ —_- if" _ o‘-_ -.