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Arithmetic Sequences and Series What is a Sequence?A Sequence is a set of things (usually numbers) that are in order. Infinite or FiniteIf the sequence goes on forever it is called an infinite sequence, otherwise it is a finite sequence Examples: {1, 2, 3, 4 ,...} is a very simple sequence (and it is an infinite sequence) {20, 25, 30, 35, ...} is also an infinite sequence {1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite sequence)
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{4, 3, 2, 1} is 4 to 1 backwards {1, 2, 4, 8, 16, 32, ...} is an infinite sequence where every term doubles{a, b, c, d, e} is the sequence of the first 5 letters alphabetically {f, r, e, d} is the sequence of letters in the name "fred"{0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s (yes they are in order, it is an alternating order in this case) In Order When we say the terms are "in order", we are free to define what order that is! They could go forwards, backwards ... or they could alternate ... or any type of order you want! Like a Set A Sequence is like a Set, except: the terms are in order (with Sets the order does not matter) the same value can appear many times (only once in Sets) Example: {0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s. The set would be just {0,1} Notation Sequences also use the same notation as sets: {3, 5, 7, ..} list each element, separated by a comma,
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and then put curly brackets around the whole thing. The curly brackets { } are sometimes called "set brackets" or "braces". A RuleA Sequence usually has a Rule, which is a way to find the value of each term. Example: the sequence {3, 5, 7, 9, ...} starts at 3 and jumps 2 every time: As a FormulaSaying "starts at 3 and jumps 2 every time" is fine, but it doesnt help us calculate the: 10th term, 100th term, or nth term, where n could be any term number we want. So, we want a formula with "n" in it (where n is any term number). So, What Would A Rule For {3, 5, 7, 9, ...} Be?
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Firstly, we can see the sequence goes up 2 every time, so we canguess that a Rule will be something like "2 times n" (where "n" is the term number). Lets test it out: Test Rule: 2n n Term Test Rule 1 3 2n = 2×1 = 2 2 5 2n = 2×2 = 4 3 7 2n = 2×3 = 6That nearly worked ... but it is too low by 1 every time, so let us try changing it to: Test Rule: 2n+1 n Term Test Rule 1 3 2n+1 = 2×1 + 1 = 3 2 5 2n+1 = 2×2 + 1 = 5 3 7 2n+1 = 2×3 + 1 = 7 That Works! So instead of saying "starts at 3 and jumps 2 every time" we write this: 2n+1 Now we can calculate, for example, the 100th term: 2 × 100 + 1 = 201 Many Rules
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But mathematics is so powerful we can find more than one Rule that works for any sequence. Example: the sequence {3, 5, 7, 9, ...} We have just shown a Rule for {3, 5, 7, 9, ...} is: 2n+1 And so we get: {3, 5, 7, 9, 11, 13, ...} But can we find another rule? How about "odd numbers without a 1 in them": And we would get: {3, 5, 7, 9, 23, 25, ...} A completely different sequence!And we could find more rules that match {3, 5, 7, 9, ...}. Really we could.So it is best to say "A Rule" rather than "The Rule" (unless you know it is the right Rule). Notation To make it easier to use rules, we often use this special style: xn is the term n is the term number Example: to mention the "5th term" you just write: x5
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So a rule for {3, 5, 7, 9, ...} can be written as an equation like this: xn = 2n+1 And to calculate the 10th term we can write: x10 = 2n+1 = 2×10+1 = 21 Can you calculate x50 (the 50th term) doing this? Here is another example: Example: Calculate the first 4 terms of this sequence: {an} = { (-1/n)n } Calculations: a1 = (-1/1)1 = -1 a2 = (-1/2)2 = 1/4 a3 = (-1/3)3 = -1/27 a4 = (-1/4)4 = 1/256 Answer: {an} = { -1, 1/4, -1/27, 1/256, ... } Fibonacci Sequence This is the Fibonacci Sequence0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...
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The next number is found by adding the two numbers before it together: The 2 is found by adding the two numbers before it (1+1) The 21 is found by adding the two numbers before it (8+13) etc... Rule is xn = xn-1 + xn-2 That rule is interesting because it depends on the values of the previous two terms. Rules like that are called recursive formulas.The Fibonacci Sequence is numbered from 0 onwards like this:n = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...xn = 0 1 1 2 3 5 8 13 21 34 55 89 144233377 ... Example: term "6" would be calculated like this: x6 = x6-1 + x6-2 = x5 + x4 = 5 + 3 = 8 Arithmetic Sequences In an Arithmetic Sequence the difference between one term and the next is a constant. In other words, you just add some value each time ... on to infinity.
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Example: 1, 4, 7, 10, 13, 16, 19, 22, 25, ... This sequence has a difference of 3 between each number. Its Rule is xn = 3n-2 In General you could write an arithmetic sequence like this: {a, a+d, a+2d, a+3d, ... } where:a is the first term, and d is the difference between the terms(called the "common difference") And you can make the ruleby: xn = a + d(n-1)(We use "n-1" because d is not used in the 1st termExercise: Find the next term and the general formula for thefollowing: A.{2, 5, 8, 11, 14, ...} B. {0, 4, 8, 12, 16, ...} C. {2, -1, -4, -7, -10, ...}For each of these three sequences there is a common difference. In the first sequence the common difference is d = 3, in thesecond sequence the common difference is d = 4, and on thethird sequence the common difference is d = -3. We will call a
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sequence an arithmetic sequence if there is a common difference. The general formula for an arithmetic sequence is an = a1 + (n - 1)dExampleWhat is the difference between the fourth and the tenth terms of{2,6,10,14,...) ,We have a10 - a4 = (10 - 4)d = 6(4) = 24 Arithmetic Series First we see that 1+ 2 + 3 + ... + 100 = 101 + 101 + ... + 101 (50 times) =101(50) In general 1 + 2 + 3 + ... + n =Example : What is S = 1 + 4 + 7 + 10 + 13 +... + 46Solution :S = 1 + (1 + 1(3)) + (1 + 2(3)) + (1 + 3(3)) + ... + (1+ 15(3)) = (1 + 1 + ... + 1) + 3(1 + 2 + 3 + ... + 15)= 16 +3(15)(16)/2 Or Alternatively Sn = n/2(a1 + an)ExampleHow much will I receive over my 35 year career if my startingsalary is $40,000, and I receive a 1,000 salary raise for each yearI work here?Solution
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We have the series: 40,000 + 41,000 + 42,000 + ... + 74,000 = 35/2 (40,000 + 74,000) = $1,995,500 An important result : if a,b,c are in A.P. then b – a = c – b ⇨ 2b = a+c , d = and nth term = Tn = Sn – Sn-1 Arithmetic meam between two numbers a & b is (a+b)/2 is denoted by A.M.A.P.Question:3 In an A.P., the first term is 2 and the sum of the firstfive terms is one-fourth of the next five terms.Show that 20thterm is -112. Answer: S5 = ¼ (S10 - S5) ⇨ 5 S5 = S10 ⇨ 5.5/2 [2a+(n- 1)d] = 10/2[2a+(n-1)d] ⇨ d = -6 ∵a=2 a20 = a+19d = 2+19 (-6) = -112.Question: 5 In an A.P., if pth term is 1/q and qth term is 1/p,prove that the sum of first pq terms is ½ (pq+1), where p≠q. Answer: pth term = a+(p-1)d = 1/q………..(1) qth term= a+(q-1)d = 1/p……….(2)
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By subtracting (1)&(2) , we get d(p-1-q+1) = - ⇨ d= by putting the value of d In (1) a = ⇨ Spq = pq/2 [ 2 +(pq- 1) ] = ½ (pq+1). Question: 9 The sum of first n terms of two arithmeticprogressions are in the ratio . Find the ratio of their 18thterm. Answer: = = = ……(1) ⇨ = By putting n=35 in (1) = = . Question:11 Sum of first p, q, r terms of an A.P. are a,b,crespectively. Prove that (q - r) + (r - p) + (p – q) = 0.Answer: Sp = p/2 [2a1 + (p – 1)d] = a Sq = q/2[2a1 + (q – 1)d] = b
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Sr = r/2[2a1 + (r – 1)d ] = c 1/2 [2a1 + (p – 1)d] (q – r) = (q - r)………………(i) 1/2[2a1 + (q – 1)d] (r – p) = (r – p)………………(ii) 1/2[2a1 + (r – 1)d ] (p – q) = (p – q)………………(iii) (i) + (ii) + (iii) (q - r) + (r - p) + (p – q) = ½{2a1 (q – r + r – p + p - q) + d [ (p-1)(q-r) + (q-1)(r-p) + (r-1)(p-q) ]} = ½ { 0 + d [ pq – q –pr +r + qr – r – pq +p + rp –p – qr +q]} = 0Question:15 If is the A.M. between a and b, thenfind the value of n. [Hint : = ⇨ 2 = (a+b)( )⇨ 2an +2 bn = an+abn-1+ban-1+bn an-1(a-b) - bn-1(a-b) = 0 ⇨(a-b)(an-1 - bn-1) = 0 ⇨ (a-b)=0 or(an-1 - bn-1) =0 ⇨ n=1 ∵ a≠b.
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Question: 16 Between 1 and 31 , m numbers have been insertedin such a way that the resulting sequence. In an A.P. and ratio of7th and (m – 1)th numbers is 5: 9.Answer: Let a2, a3, a4,…………am+1 as 1, a2, a3,a4,…………am+1, 31am+2 = a1+ (m+1) d = 31 and a1 = 131 = 1 + (m+1) d ⇨ d = 30/(m+1) or d =(b – a)/ (n+1) According to statement = ⇨ = = ⇨ m = 14.Geometric Sequences In a Geometric Sequence each term is found by multiplying the previous term by a constant. Example: 2, 4, 8, 16, 32, 64, 128, 256, ... This sequence has a factor of 2 between each number. Its Rule is xn = 2n In General you could write a geometric sequence like this: {a, ar, ar2, ar3, ... } where:
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a is the first term, and r is the factor between the terms (called the"common ratio") r can be calculated by dividing any two consecutiveterms in a geometric sequence. The formula for calculatingr is... ...where n is any positive integer greater than 1. Note: r should not be 0. When r=0, you get the sequence {a,0,0,...} which is not geometric And the rule is: xn = ar(n-1) (We use "n-1" because ar0 is the 1st term)Finding the Sum of a SeriesGiven our generic arithmetic sequence a1, a2, a3, a4, ...,an, we can look at it as a series: a1 + a2 + a3 + a4 + ... +an. There exists a formula that can add a finite list ofthese numbers and a formula for an infinite list of these
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numbers. Here are the formulas... and...where Sn is the sum of the first n numbers, a1 is thefirst number in the sequence, r is the common ratio ofthe sequence, and -1 < r < 1 for infinite series.***Introduction The process of adding infinitely manynumbers is at the heart of the mathematical concept of anumerical series. Let ‘s denote the sum of the series just considered: Lets multiply both sides by 1/2 and subtract the second line from the first. All terms on the right side except for the 1 will cancel out!
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We have shown thatOne also says that this series converges to 2.Lets play the same game for a general q instead of 1/2:multiply both sides by qthen, subtract the second line from the first:The seriesis called the geometric series. It is the most importantseries you will encounter!S∞ = , a is first term , r is the common ratio.
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As n→∞ rn → 0 for |r|<1. Geometric mean between two numbers a & b is i.e.,G2 = ab or G = , Harmonic mean (H.M.)= If a , b, c are in G.P. then b/a = c/b ⇨ b2 = ac. A – G = - ≥0⇨ A ≥ G≥H(Harmonic mean) R= Special sequences: (i) 1+2+3+…………..+n = (ii) 12+22+32+……..+ n2 = (iii) 13+23+33+………+n3 = ( )2Example:Find the sum of the seriesFirst, factor out the 5 from upstairs and a 2 from downstairs:
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.The series in the parentheses is the geometric series with ,but the first term, the "1" at the beginning is omitted.Thus, the series sums up toThere is a slightly slicker way to do this. Do you see how?Try it yourself!Find the sum of the seriesG.P.Question: 21 Find four numbers forming a G.P. in whichthe third term is greater than the first term by 9, and thesecond term is greater than the 4th term by 18.Answer: In G.P. a, ar, ar2, ar3 are first four terms.
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ar2 – a = 9 and ar3 - ar = 18 , dividing by above results , we get r =-2 and a=3, then termsare 3, -6, 12 and -24.Question:23 If the first and the nth term of a G.P. are a and bresp.and if P is the product of n terms , prove that P2 = (ab)n .Answer: P = a.ar. ar2. ar3……..arn-1 = an. r(1+2+3+……+n-1) =an . [+2+3=……+n=n(n+1)/2] P2 = ( an . )2 = a2n . = [a.(a rn-1]n =(ab)n ∵ b = nth term = a rn-1Question:24 Show that the ratio of the sum of first nterms of a G.P. to the sum of terms from (n+1)th to (2n)th Termis .Answer: Sn = ……….(i) The sum of terms from (n+1)th to (2n)th = ……(ii)[ a,ar,ar2,…….arn-1, arn,arn+1, ……ar2n-1]Question:28 The sum of two numbers is 6 times their G.P.,show that numbers are in the ratio (3+2 3-2 ).
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Answer: = by c & d = ⇨ = , again by c& d,we get = , then squaring on the both sides ,we getthe answer.(MISC.)Question:4 Find the sum of all numbers between 200 and 400which are divisible by 7.Answer: 203, 210, 217, 224, ……. ,339. 399 = 203 + (n-1)7 ⇨ 196 = (n-1)7 ⇨ n = 29, so Sn =n/2 (a+l) = 29/2(203=399) =29 301 = 8729.Question: 5 Find the sum of integers from 1 to 100 that aredivisible by 2 or 5. Answer: Divisible by 2 are 2,4,6,……..,94,96,98,100 ordivisible by 5 are 5,10,15…….95,100. nth term = 100 = 2+(n-1) 2 ⇨ n = 5010 terms as 10,20,30,……100 of 5,10,15…….95 are alreadyused in 2,4,6,…….100 (100-40[50-10]=60) The sum of remaining 60 terms is 60/2[2+100] = 3060.Question:11 A G.P. consists of an even number of terms. If thesum of all the terms is 5 times the sum
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Of terms occupying odd places, then find the common ratio.Answer: a, ar, ar2, ar3 ,…………arn-1 in G.P. S2n = 5[ a+ ar2 +ar4+……n terms] = 5. ⇨ 1= ⇨ r = 4.Question: 18 If a and b are the roots of x2 – 3x +p = 0 and c, dare the roots of x2 – 12x + q = 0,where a,b,c,d form an G.P. Prove that (q+p) : (q-p) = 17:15.Answer: a+b = 3, ab = p and c+d = 12, cd = qa, b, c, d are in G.P. i.e., b=ar, c = ar2 d= ar3a+b= a+ar=a(1+r) = 3 …..(i) and c+d=12 ⇨ ar2 + ar3 = ar2(1+r)= 12….(ii) , by dividing (i)&(ii) 1/r2 = ¼ ⇨ r = ±2 , then = = =17:15.Question:20 If a, b, c are in A.P. ; b, c, d are in G.P. and 1/c, 1/d,1/e are in A.P. Prove that a, c, e are in G.P.Answer: 2b =a+c…..(i) c2 =bd…..(ii) 2/d = 1/c+1/e = ⇨ d= ........(iii)We shall prove c2 = aeFrom (i),(ii) &(iii) we get,
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c2 = )( = ⇨ c2 (c+e) = ace+c2e ⇨ c3 + c2e =ace+c2e ⇨ c2 = ae.Question:23 Find the sum of 3+7+13+21+31+.........Answer: an = n(n+1)+1 ,n=1,2,3…. = n2+n+1 Sn = + +n = + +n.Question:27 A farmer buys a used tractor for Rs. 12000. Hepays Rs. 6000 cash and agrees to pay the balance in annualinstalments of Rs. 500 plus 12% interest on the unpaid amount.How much will the tractor cost him?Answer: Total cost =Rs. 12000 , paid cash = 6000 then balance= Rs. 6000 No. of instalments of Rs. 500 = 12Interest on 1st instalment = Rs. =Rs.720 1st instalment = Rs.(500+720)=1220Interest on 2nd instalment = Rs. =Rs.6602nd instalment = Rs.(550+660)=1160 and so on… Total amount in instalments = (1220+1160+1100+……to 12terms)
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S= 12/2[2 1220 + (12-1)(-60)] = 10680 , therefore totalpaid by farmer = Rs. 6000+10680=16680.Question: 32 150 workers were engaged to finish a job in acertain number of days. 4 workers dropped Out on second day, 4 more workers dropped out on third dayand so on.It took 8 more days to finish the work. Find thenumber of days in which the work was completed. Answer: First term is 150 common difference is -4 (4 workersdropped every day) n/2[2 150+(n-1) -4] = n(152-2n)Had the workers not dropped out , then the work would havefinished in (n-8) days with 150 workers working on each day .The total no. of workers who would have worked all then daysare 150(n-8)n(152-2n) = 150(n-8) ⇨ n=25 ∵ n≠-24.ASSIGNMENT OF SEQUENCE & SERIESQuestion:1 Find k so that 2/3, k, 5k/8 are in A.P.Question:2 If the roots of (b-c)x2+(c-a)x+(a-b) = 0 are equal,then a,b,c are in A.P. [Hint: take Discriminant D=0]
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Question:3 There are n A.M.’s between 7 and 85 such that (n-3)th mean : nth mean is 11 : 24.Find n.[Hint: 7, a2, a3, a4,…………..an+1, 85 ⇨ an+2 = 85 ⇨ d= sameas in your ncert book problem, find n=5] Question:4 Prove that the sum of n terms of the series 11+103+1005+ ……..is ( - 1) + n2[HINT: Sn = (10+1)+(100+3)+ (1000+5)+……….n terms = (10+102+103+……n term)+(1+3+5+……n terms) G.P. A.P.Question:5 How many terms of the series 2+2 +4+…..willamount to 30+14 .[Hint: use Sn = = 30+14 ,a=2 & r = , then n=7]Question:6 Find the sum of (1+ ) + ( + )+( )+…….to ∞[Hint: (1 + + +…..) + ( ), answer is 7/3 ,use S∞= , a is first term , r is the common ratio.]Question:7 Show that (x2+xy+y2) , (z2+zx+x2), (y2+zy+z2)are consecutive terms of A.P. if x, y, z are in A.P.
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[Hint: let a = (x2+xy+y2) , b = (z2+zx+x2) ,c = (y2+zy+z2)use b-a = c-b , then you will getif (x+z)2 – y2 = y(x+y+z) ⇨ x+z = 2y]Question:8 If a, b, c, d are in G.P. prove that a2-b2, b2-c2, c2-d2are also in G.P. [HINT: take a, b= ar, c=ar2, d = ar3 , show (b2-c2)/ (a2-b2)=(c2-d2)/ (b2-c2) by putting values of a, b, c, d]Question:9 If one geometric mean G and two arithmeticmean p and q be inserted between two quantities Show that G2 = (2p-q)(2q-p).[Hint: G2 = ab, a, p, q, b are in A.P.d(common difference)= (b-a)/3 . p = a+d= , q = a+2d = put these values inR.H.S.]Question:10 If the value of 1+2+3+…….+n = 28, then findthe value of 12+22+32+………..+n2 .**Question:11 If g1, g2 be two G.M.’s between a and b and Ais the A.M. between a & b, then prove that + = 2A.
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[Hint: 2A = a+b, b/g2 = g2/g1 = g1/a ⇨ a = g12 / g2 , b =g22 /g1.]**Question:12 If a is the A.M. of b, c and two geometricmeans between b , c and G1,G2, then prove that G13 = g23 .Answer: 2a = b+c , c = ar3 ⇨ r = , put in G1= br &G2= br2 .**Question:13 If x = 1+a+a2+a3+……….∞ and y =1+b+b2+b3+………. ∞ Then prove that 1+ab+a2b2+….. ∞ =Answer: x = 1+a+a2+a3+……….∞ = y = 1+b+b2+b3+………. ∞ = take R.H.S. = = (1-ab)-1 = 1+ab+a2b2+….. ∞ [by binomial expansion (1-x)-1 = (1+x+x2+x3+…..)]**Question:14 An A.P. consists of n(odd)terms and it’smiddle term is m. Prove that Sn = mn. Answer: m = mid term = T(n+1)/2 = a+( - 1)d ⇨ 2m=2a+(n-1)d , Sn = n/2[2a+(n-1)d]=mn.**Question:15 Sum of infinity the series
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+ + +………..[Hint: = == ⇨ S∞=1 [ + +......∞ =1 as given short-cutmethod on blog)].**Question:16 Prove that 23rd term of sequence 17, 16 , 15 ,14 ,………is the first negative term.[Hint: a=17, d=-4/5 and let nth term be first negative term ⇨17+(n-1)(-4/5) < 0 ⇨ n > 89/4 ⇨ n=23]**Question:18 Prove that 1+ + + + + ………..∞ = 3.Answer: we can write above series as 1 + [1++ +…….],where a=1,d=2 and r= 1/3, then use formula ofcombined A.P.& G.P(Arithmetic –geometric series) [ + ] or you can do by another method S - S = {1+ ++ + + ………..∞} – { + + + + + ………..∞} ⇨ = 1+ ( - ) + {1+ + +…..∞} =2 ⇨ S=3 a/(1-r)**Question: 19 If there are distinct real numbers a,b,c are inG.P. and a+b+c = bx , show that x ≤ -1 or x ≥ 3.
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[Hint: take D ≥ 0 , a+ar+ar2 = (ar)x ⇨ r2+(1-x)r+1=0]**Question:20 (i) If first term of H.P. is 1/7 and 2nd term is1/9, prove that 12th term is 1/29.[Hint: as 1/a,1/b,1/c are in H.P. ⇨ a,b,c are in A.P. thereforefirst and second terms are in A.P. will be 7,9 a=7, d= 2 , then find a12] (ii) In an increasing G.P., the sum of first and last term is66, and product of the second and last but one term is 128. Ifthe sum of the series is 126, find the number of terms in theseries.[Hint: a+arn-1 = 66 , (ar)( arn-2) = 128 and Sn = 126 ⇨ r=2and n = 6] Terms Arithmetic Sequence - A sequence in which each term is a constant amount greater or less than the previous term. In this type of sequence, a n+1 = a n + d , where d is a constant. Common Ratio - In a geometric sequence, the ratio r between each term and the previous term. Convergent Series - A series whose limit as n→∞ is a real number. Divergent Series - A series whose limit as n→∞ is either ∞ or - ∞ .
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Explicit Formula - A formula for the n th term of asequence of the form a n = some function of n .Finite Sequence - A sequence which is defined onlyfor positive integers less than or equal to a certaingiven integer.Finite Series - A series which is defined only forpositive integers less than or equal to a certain giveninteger.Geometric Sequence - A sequence in which theratio between each term and the previous term is aconstant ratio.Index of Summation - The variable in the subscriptof Σ . For a n , i is the index of summation.Infinite Sequence - A sequence which is defined forall positive integers.Infinite Series - A series which is defined for allpositive integers.Recursive Sequence - A sequence in which ageneral term is defined as a function of one or moreof the preceding terms. A sequence is typicallydefined recursively by giving the first term, and theformula for any term a n+1 after the first term.Sequence - A function which is defined for thepositive integers.Series - A sequence in which the terms aresummed, not just listed.Summation Notation - an=a1+a2+a3+a4+ ... + a n . The symbol Σ and its subscript andsuperscript are the components of summationnotation.
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Term - An element in the range of a sequence. Asequence is rarely represented by ordered pairs, butinstead by a list of its terms.