Tiêu đề XI - maths - chapter 9 - SEQUENCE_SERIES (140-173).pdf ✅

140 NARAYANAGROUP SEQUENCE & SERIES JEE-MAIN-SR-MATHS VOL-I SEQUENCE AND SERIES Sequence: A set of numbers is arranged in a definite order according to some definite rule is called a sequence. e.g. 2, 4, 6, 8, ....., is a sequence  A sequence is a function whose domain is a set of natural numbers. If the range of a sequence is a subset of real numbers (or complex numbers), then it is called a real sequence (or complex sequence) Series: The sum of the terms of a sequence is called a series.  If 1 2 3 a a a , , ,...... is a sequence, then the expression 1 2 3 a a a   ...... is a series  A series is called finite series, if it has finite number of terms. Otherwise it is called infinite series. e.g .i) 1+3+5+ ..........+21 is a finite series. ii) 2+4+6+8+....... is an infinite series.  Sequences following specific patterns are called progressions. Arithmetic progression (A.P):-  A sequence is called an arithmetic progression, if the difference between any two consecutive terms is the same.  A.P is of the form a a d a d a d , , 2 , 3    ..... where a is 1st term and d is common difference General term of an A.P:  Let ‘a’ be the first term and ‘d’ be common difference of an A.P, then its genaral term (or)   th n term is 1 T a n d n     If ‘l’ be the last term and ‘d’ be common difference of an A.P, then th m term from the end ' Tm = l-(m-1)d  th m term from the end = th (n-m+1) term from the beginning. W.E-1: Find the first negative term of the sequence 1 1 3 20,19 ,18 ,17 ,...... 4 2 4 Sol: The given sequence is an A.P in which first term a=20 and common difference d=-3/4. Let the nth term of the given A.P. be the first negative term. Then, an  0          a n d n ( 1) 0 20 ( 1)( 3/ 4) 0 83 3 0 83 3 0 3 83 4 4 n         n n 2 27 3   n   n 28 thus, 28th term of the given sequence is the first negative term. W.E-2: If 100 times the 100th term of an A.P with non-zero common difference equals the 50 times of 50th term, then find 150th term of this A.P. (AIEEE 2012) Sol: 100 50 100 50 T T  ; 100(a+99d)=50(a+49d) 2a+198d=a+49d ; a+149d=0 T a d 150    149 0 Sum to n terms of an A.P:   2 n n S a l   2 1   2 n      a n d   where a  first term, l  last term d  common difference  If the sum of n terms of a sequence Sn is given, then its th n term Tn can be determined by T S S n n n   1 W.E-3: How many terms are to be added to make the sum 52 in the series (-8)+(-6)+ (-4)+....? Sol: 52 2 8 1 2 52     2 n n S n              n n 2 18 104       n n n  9 52 13  SYNOPSIS
NARAYANAGROUP 141 JEE-MAIN-SR-MATHS VOL-I SEQUENCE & SERIES W.E-4: Let 1 2 , ,....., a a an be the terms of an A.P. If 1 2 1 2 ..... ..... p q a a a a a a       = 2 2 p q , p q  then find 6 12 a a . Sol:         2 1 1 2 1 1 1 6 21 1 6 21 2 1 2 1 2 2 1 2 1 2 1 2 , 11, 41 1 2 11 41 p a p d p p a p d q q q a q d a q d p a d p a For p q q q a a d a a                                                  Properties of A.P:-  a b c , , are in AP    2b a c  In a finite A.P, the sum of the terms equidistant from the begining and the end is always same and is equal to the sum of the first and last term 2 1 3 2 4 3 1 . ., n n n n i e a a a a a a a a            a a a a 1 2 3     ...... n      middleterm , if isodd sumof twomiddleterms , if iseven 2 n n n n          If 1 2 3 , , ........... a a a an are in A.P then a) 1 3 2 1 , ,..... , , a a a a a n n are in A.P b) 1 2 3 , , ; ....... a a a a         n are in A.P (where R ) c) 1 2 3 , , ; .......     a a a an are in A.P (where   R {0})  th p term of an A.P. is ‘q’ and th q term is ‘p’, then Tp q  0  If th m term of an A.P. is ‘n’ and th n term is ‘m’ then th p term is ‘m+n-p’  If S q p  and S p q  for an A.P., then S p q p q      W.E-5:If     1 3 3 1 1, log 3 2 ,log 4.3 1 2 x x   are in A.P, then find x. Sol:     1 3 3 1 1, log 3 2 ,log 4.3 1 2 x x   are in A.P     1 3 3 log 3 2 1 log 4.3 1      x x     1 3 3 3 log 3 2 log 3 log 4.3 1      x x     1 3 3 log 3 2 log 3 4.3 1     x x     1 3 2 3 4.3 1     x x 3.3 2 12.3 3     x x   3 2 12 3, 3x t where t t      2 2         3 2 12 3 12 5 3 0 t t t t t     4 3 3 1 0 t t     3 1 3 , 3 3 0 4 3 4 x x t        3 3 3 log 1 log 4 4 x           Selection of terms in an A.P: W.E-6: If the sum of four numbers in A.P is 24 and the sum of their squares is 164 then find those numbers. Sol: (a-3d)+(a-d)+(a+d)+(a+3d)=24     4 24 6 a a         2 2 2 2 a d a d a d a d         3 3 164     2 2 2 2      2 9 2 164 a d a d 2 2 2          a d d d 5 41 36 5 41 1 required numbers are 3,5,7,9
142 NARAYANAGROUP SEQUENCE & SERIES JEE-MAIN-SR-MATHS VOL-I Some Facts about A.P:-  If 1 2 3 , , ........ a a a an and 1 2 3 , , ........ b b b bn are two A.P’s then a) 1 1 2 2 a b a b   , ,......... are in AP b) 1 1 2 2 3 3 a b a b a b , , ......... and 1 2 3 1 2 3 , , .......... a a a b b b are not in A.P c) If the terms of an A.P. are chosen at regular intervals, then they form an A.P  If a constant ‘k’ is added to each term of A.P., with common difference ‘d’, then the resulting sequence also will be in A.P., with common difference (d+k).  If every term is multiplied by a constant ‘k’, then the resulting sequence will also be in A.P., with the first term ‘ka’ and common difference ‘kd’.  If th n term of the sequence T An B n   (i.e) [Linear expression in n ] then the sequence is A.P with first term is ‘A+B’ and common difference A coefficient of n    If sum of n terms of a sequence is 2 S An Bn C n    (i.e.Quadratic exprssion in n) then the sequence is A.P with first term is 3A+B and common difference is 2A. Also in this sequence th n term Tn = 2An + (A+B)  If the ratio of the sums of n terms of two A.P.’s is given then the ratio of their th n terms may be obtained by replacing n with 2 1 n   in the given ratio.  If the ratio of th n terms of two A.P.’s is given, then the ratio of the sums of their n terms may be obtained by replacing n with 1 2 n  in the given ratio  Sum of the interior angles of a polygon of ‘n’ sides is   0 n  2 180  The th n common term of two Arithmetic Series is ( L.C.M of common difference of 1st series and 2nd series )( n-1)+ 1st common term of both series W.E-7: Find the th n term of the sequence 5,15,29,47,69,95,... Sol: The given sequence is not an A.P. but the successive differences between the various terms i.e. (15-5),(29-15),(47-29),(69-47),(95-69),.... i.e. 10,14,18,22,26,..... are in A.P Let th n term of the given sequence be   2 t an bn c n     1 Putting n=1,2,3 in 1 , we get t a b c a b c 1         5 2  t a b c a b c 2         4 2 4 2 15 3  t a b c a b c 3         9 3 9 3 29 4  Solving (2),(3),(4), we get a=2,b=4,c=-1.  the th n term of the given sequence is 2 t n n n    2 4 1 W.E-8:The sum of the first n terms of two A.P’s are in the ratio (2n+3):(3n-1). Find the ratio of 5th terms of these A.P’s. Sol: Given that ' 2 3 3 1 n n S n S n    The ratio of nth terms     ' 2 2 1 3 4 1 3 2 1 1 6 4 n n t n n t n n         ' 5 5 t t: 21: 26  W.E-9: The interior angles of a polygon are in A.P. the smallest angle is 1200 and the common difference is 0 5 . Find the number of sides of the polygon . Sol: Given a=1200 , d= 0 5 Sum of the interior angles of a polygon of n sides is   0 n  2 180 2 120 1 5 2 180       2 n        n n       n n n 5 235 2 360        5 47 2 360 n n n       2     n n n 47 2 72    2         n n n n 25 144 0 9 16 0   n or 9 16 (Since neglecting n=16, Since that case largest angle is [120+(15)5]=195, which is not possible no longer angle of a polygon is more than 180) n=9
NARAYANAGROUP 143 JEE-MAIN-SR-MATHS VOL-I SEQUENCE & SERIES W.E-10: Find 12th common term of two Arithmetic Series 7+10+13+..... and 4+11+18+......... . Sol: The th n common term of between two series = ( L.C.M of common difference of 1st series and 2nd series )( n-1) + 1st common term of both series. =(L.C.M of 3,7) (12-1)+25 =21(11)+25 =256 W.E-11: Find the number of common terms to the two sequences 17,21,25,...,417 and 16,21,26,...,466. Sol: series 17,21,25,.,417 has common difference4 series 16,21,26,...,466 has common difference 5 LCM of 4 and 5 is 20, the first common term is 21. Hence, the series is 21,41,61,...,401; which has 20 terms. Arithmetic mean (A.M): The Arithmetic mean A of any two numbers a and b is given by 2 a b  , where a A b , , are in AP  If 1 2 3 , , ...... a a a an are n numbers then Arithmetic mean A of these numbers is given by  1 2  1 ..... A a a an n      The n numbers 1 2 3 , , ....... A A A An are said to be Arithmetic means between a and b if 1 2 3 , , , ......... , a A A A A b n are in AP Here a  First term b n th    2 term =a+(n+1)d then, 1 b a d n    1 1 b a A a n       2 2 1 b a A a n     , .....   1 n n b a A a n     1 2 3 ..... 2 n a b A A A A n             W.E-12 : If n arithmetic means are inserted between 2 and 38, then the sum of the resulting series is obatined as 200, then find the value of n. Sol: We have   2 2 38 200 2 10 8 2 n n n         Arithmetic mean of the mth power : Let 1 2 , ,... a a an be n positive real number (not all equal) & let m be real number then 1 2 ..... m m m a a an n      1 2 ..... 0,1 m n a a a m R n                  1 2 1 2 ... 0,1 ... 0,1 m n m n a a a m n a a a m n                        W.E-13: Prove that 1 1 2 ..... 2 n n n      Sol: 1 1 2 ..... 1 2 3 ... n n 2 n n                   1 2 1 2 1 1 2 2 n n n n                       1 1 2 ..... 2 n n n       Geometric Progression (G.P):- A Sequence is called a Geometric progression, if the ratio of any two consecutive terms is the same  G.P is of the form 2 3 a ar ar ar , , , ........, Where a is the first term and r is the common ratio Genaral term of G.P:- If ‘a’ be the first term and ‘r’ be the common ratio, then general term (or) th n term of G..P is n 1 T ar n    The th n term from the end of a finite G.P consisting of m terms = m n ar 