CBSE, ClassXI

MADE BY SANJEEVSequences and Series

SequenceA sequence is an ORDERED list of objects Different from sets (in sets order does not matter)The set A = {1,2,3,4,5} is equal to B = {5,4,3,2,1} because each element of A B and each of B A The sequences S = 1,2,3,4,5 and T = 5,4,3,2,1 are different because ordering matters

Ordering=> we can refer to nth element of a SequenceWe can say 5th element of a sequence of integersElement can occur multiple times unlike for a set

Examples of SequencesA finite sequencehas finite number of terms: e.g. 5,4,3,2,1 If a sequence is not finite it is called an infinite sequence -1,1,-1,1,-1,1,-1,... first element is -1, second is 1, and nth element is -1 to the power n 3,6,9,12,15,18... The nth element is 3*n 2,4,8,16,32,... The nth element is 2 to the power n

Arithmetic ProgressionsA sequencewhere, other than first term every term = previous term + common difference

Notation:a = first term L = last term d = common difference S(n) = sum to n terms (Note: Slightly different notation from NCERT text)

General Term of Arithmetic Progression1st Term = a = a + (1-1)*d 2nd Term = a + d = a + (2-1)*d 3rd Term = a + 2d = a + (3-1)*d kth Term = a + (k-1)*d If the series has n terms, then Last term is nth term and so last term = L = a + (n-1)*d

Example:First + Last Term = 80 for an AP with large no. of termsWhat is (a) Sum of 2nd term + last but 1 term (b) Sum of 3rd term + last but 2 term

2nd Term = First Term + common diff.Last but 1 Term = Last Term common diff.2nd + Last but 1 Term = First Term + Last Term

3rd Term = 2nd Term + common diff.Last but 2 Term = Last but 1 Term common diff.3rd + Last but 2 Term = 2nd Term + Last but 1 Term = First Term + Last Term

So answer must be 80 for both questions.

Summing Arithmetic ProgressionsAP has the propertyfirst term + last term=2ndterm + last but 1 term=3rd term + 2nd before last term= a + (a+(n-1)*d) = (a+L)normal sum = first+ second + ... + lastreverse sum = last+ last but 1 + ... + firstnormal + reverse = (first + last) + (second + last but one) + ... + (last + first)normal + reverse = (first + last term)*(number of terms)normal sum*2 = (first + last term)*(number of terms)normal sum = (first + last term)*(number of terms)/2

Example: A.N. Kolmogorov at Age 5Worked outsum of first n odd numbers for school journalfirst odd number = 1 = first term2nd odd number = 3 = 1 + (2-1)*2nth odd number = 1 + (n-1)*2 = last term 1,3,5,7,...till 1+(n-1)*2 is an AP with common difference 2 Summing this series yields an interesting answer Sum = [first term + last term]*(number of terms)/2 [1 + {1+(n-1)*2}]*n/2 = [1+1+2n-2]*n/2 = n*n 1+3 = 2*2; 1+3+5=3*3; 1+3+5+7=16=4*4;1+3+5+7+9=25=5*5

Example: Average of an APThe first term of AP is 11 andlast is 56. We are not given( a )common difference( b ) number of termsCan we find the average? YesSum of AP = [Average of AP ]*{number of terms}Sum of AP = [first + last term]*{number of terms}/2Average of AP = [first + last term]/2(True For ANY AP) So any AP starting with 11 and ending with 56 has average = (11+56)/2 = 33.5. This includesAP1 = 11,16,21,26,31,36,41,46,51,56AP2 = 11,56AP3 = 11,12,13,....56

Example: 3 Consecutive terms of APLets call the terms left, middle and right. Call the common difference d Since they are in AP we know middle = left+ dor left = middle d right =middle+d left + right = 2*middle left+middle+right = 3*middle

Example: Diff. of 2 terms of an APkth term = a + (k-1)*djth term = a + (j -1)*dkth term - jth term = (k j)*d depends only on d and (k-j) it does NOT depend on k and j but only on the difference k-jfirst term (a)15th term - 10th term = (15 -10)*d = 5d 10th term - 5th term = (10 - 5)*d = 5d 7th term - 2nd term = ( 7 - 2)*d = 5d If 18th term - 10th term is 45, what is 20th term - 4th term? 90. (20-4)d = 16d = 2*(18-10)d=2*45

Arithmetic Mean (AM) of 2 NumbersGiven two numbers, the AM = 1/2 the sum Example: Given 11,17 Arithmetic Mean = (11+17)/2 = 14. A.M. has some interesting propertiesIf the two numbers are equal, they are also equal to AMIn case numbers are differentAM is greater than smaller of the two numbersAM is lesser than larger of the two numbersAM - smaller number = larger number - AM = constantSmaller number, AM, larger number are in AP with common difference = constant in above equation

Create AP between 2 termsGiven first and last term, insertn terms to form APAP = first term, first term + d, first term + 2d, .... last term Figure outthe common difference dWe know[last term - first term]n terms between first and last meanslast term is n+2th In previous e.g., {n+2th term - first term} = (n+2-1)*d d = [last term - first term]/(n+1) Insert 9 terms between 2 and 122 so that they are in AP d = [122 - 2]/(9+1) = 120/10 = 12AP is 2,14,26,38,50,62,74,86,98,110,122

Adding a constant to each term of APCall kth term of old AP, with common difference d, old(k)old(k) = old(k-1) + d or old(k)-old(k-1) = d Call kth term of new series as new(k) Given thatnew(k) = old(k) + constant But is new series an AP? Yes, becausenew(k) = old(k) + constantnew(k-1) = old(k-1) + constantnew(k)-new(k-1) = old(k)-old(k-1) = dnew(k) = new(k-1)+dnew is AP with same common difference d as old AP

Multiplying a const. to each term of APCall kth term of old AP, with common difference d, old(k)old(k) = old(k-1) + d or old(k)-old(k-1) = d Call kth term of new series as new(k) Given thatnew(k) = old(k) * constant But is new series an AP? Yes, becausenew(k) = old(k) * constantnew(k-1) = old(k-1) * constantnew(k)-new(k-1) = [old(k)-old(k-1)]*constant = d*constant new(k) = new(k-1)+{d * constant}new is AP with common difference=d*constant

Geometric Progressions (GP)AP: any term = previous term+common difference (d)GP: any term = previous term*common ratio (r) Above defines AP and GP for all terms except the first AP: Kth term = (k-1)th term + dGP: Kth term = (k-1)th term * r Examples of GP: 1,2,4,8,16,32,64 Common ratio (r) = 21,1/2,1/4,1/8,1/16/,1/32 Common ratio (r) = 1/2

Nth term of GPA bank pays annual interest of i% on the entire balance Deposit principal P at beginning of year and no withdrawals What is his balance at beginning of each year?1st year = P 2nd year = P*(1+i) 3rd year = P*(1+i)*(1+i) 4th year = P*(1+i)*(1+i)*(1+i) = P*[(1+i) ]3 nth year = P*[(1+i) ]n-1 For a GP beginning with a and common ratio rnth term = a * [ r ]n-1 Note: nth term = (n-1)th term * r

Sum to n terms of GP

If r is 1, every term = first term,sum = n * first term = n*a S(n) =(a) +(a*r)+(a*r*r)+ ...(a*rn-1) S(n)*r = (a*r)+(a*r*r)+ ...(a*rn-1) + (a*rn) every term in RHS of S(n) is in S(n)*r except (a)S(n)*r has additional term (a*rn)S(n)*r = S(n)-a + (a*rn)S(n)*[r-1] = a*[ rn - 1 ] = [last term* r first] S(n) = {a/(r-1)} * [ rn - 1 ] Doesnt depend on nAlthough Last term a*rn-1 Sum depends on rn= [1/(r-1)]*[last term * r first]

Example: Summing a GP and Nth Term1+2+4+8+... to n terms. This GP has interesting propertiesS(n) = 1/(r-1)*[(last term * r) first term] r = common ratio = 2So (1/(r-1)) = 1first term = 1

S(n) = {(last term * 2)-1 }

1+2 = ( 2*2) 1 = 31+2+4 = ( 4*2) 1 = 71+2+4+8 = ( 8*2) 1 = 151+2+4+8+16 = ( 16*2) - 1 = 311+2++512 = (512*2) 1 = 1023

Example: 3 Consecutive terms of GPCall these the left term, middle term, and right termmiddle term = left term * r left term = middle term / rright term = middle term * r left term * right term = middle term * middle term

Product of these three terms isP = (left term)*(middle term)*(right term)P = cube of the middle term Also, (middle term/left term) = (right term/middle term) = r

Example: Ratio of any 2 terms of GPkth term = a * rkjth term = a * rjkth term/jth term = r (k-j) Observe that the ratio depends onrk jRatio does not depend onthe first termaindividual values of k, j but depends only onk-j So for example, for any GP, ratio of18th Term/7th term = r1127th Term/16th term = r11

Create GP between 2 termsGiven first and last term, insertn terms to form GPGP = first term, first term*r, first term*r*r, .... last term Figure outthe common difference ratio rWe know[last term - first term]n terms between first and last meanslast term is n+2th In previous e.g., {n+2th term/first term} = r (n+2-1) r = [last term/first term] (1/[n+1]) Insert 3 terms between 3 and 243 so that they are in GP r = [243/3] (1/[3+1]) = 81 (1/4) = 3GP is 3,9,27,81,243

Geometric Mean (GM) of 2 NumbersGM is square root of the product of the two given numbersExample: Given 4,9 GM = square root of 36 = 6 G.M. has some interesting propertiesIf the two numbers are equal, they are also equal to GMIn case numbers are different, call them Big, SmallGM = sqrt(Big*Small)GM/Small = sqrt(Big*Small)/Small = sqrt(Big/Small) > 1GMisgreaterthanSmall Big/GM= Big/sqrt(Big*Small)/ = sqrt(Big/Small) > 1GMislesser thanBig (GM/Smallnumber) =(Bignumber/GM)Smallnumber,GM,Bignumberare in GP

Relationship Between GM and AMTwo numberssand L,both >= 0, AM = (s+L)/2,GM = sqrt(s*L) If both numbers are equal AM = GM = s = LIf not lets call the larger of the two numbers LIf s is really small, say s = 0, GM = 0 but AP = L/2 > 0So we guess that AM >= GM AM-GM = (s+L)/2 - sqrt(s*L) = [s + L - 2*sqrt(s*L))] /2 = [sqrt(L) - sqrt(s)] /2 The RHS of last equation is >=0 so AM-GM >=0