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Sequences and Series

Toolkit

Arithmetic Sequences

Arithmetic sequences are sequences with a common difference, that is to say, that the difference between consecutive terms is constant.

Description Formula
General kth term tk=a+(k1)d, where a is the first term and d is the common difference
Sum of n terms Sn=n2(a+tn)=n2(2a+(n1)d)
Spacing of terms Because there is a common difference between consecutive terms we have tk+tl=tm+tn if and only if k+l=m+n

Geometric Sequences

Geometric sequences are sequences with a common ratio, that is to say, the ratio of consecutive terms is constant.

Description Formula
General kth term tk=ark1, where a is the first term and r is the common ratio
Sum of n terms Sn=a(1rn)(1r)
Spacing of terms If a0 and r0, then because there is a common ratio between consecutive terms, we have tktl=tmtn if and only if k+l=m+n.
Infinite sum If the ratio r satisfies the condition |r|<1, we can calculate the infinite sum a+ar+ar2+ar3+ using S=a1r.

Other

Arithmetic and geometric sequences are a small subset of all sequences, even though they are emphasized in high school mathematics. The following are some extensions that frequently appear on contests.

Description Formula
Sum of the first n integers k=1nk=n(n+1)2
Sum of the first n squares k=1nk2=n(n+1)(2n+1)6
Sum of the first n cubes k=1nk3=(n(n+1)2)2
Telescoping series If tk=ukuk1, then k=1ntk=k=1n(ukuk1)=unu0

Sample Problems

  1. What is the sum of all multiples of 7 or 11 less than 1000?

    Solution

    Since we are adding (7+14+21+28++994)+(11+22+33++990), we are adding two arithmetic sequences. However the multiples of 77 are included in both sequences and so must be subtracted (in order to avoid counting them twice) after we add the two sequences above. Therefore, the required sum is (7+14+21+28++994)+(11+22+33++990)(77+154++924).

    The term 994 is the 142nd term in the first sequence and so the sum of the first sequence is 1422(7+994). The term 990 is the 90th term in the second sequence and so the sum of the second sequence is 902(11+990). The term 924 is the 12th term in the sequence of terms we remove and so the sum of that sequence is 122(77+924). Thus, the required sum is 1422(7+994)+902(11+990)122(77+924)=(71+456)(1001)=(110)(1001)=110110

  2. A sequence is given such that t1=1 and tn+1=tn+3n2+3n+1. Evaluate t100.

    Solution

    Since the difference, tntn1 is not constant, the series is not arithmetic. Setting n=1, we find t2=1+3+3+1=8 Setting n=2, we find t3=8+12+6+1=27 These facts suggest tn=n3 for every n.

    To prove that tn=n3 is an alternate definition for the same sequence, we first note that t1=1=13. Further, consider two adjacent terms in the sequence given by the alternate definition, i.e. tn=n3 and tn+1=(n+1)3. Then the difference between these terms is tn+1tn=(n+1)3(n)3=(n3+3n2+3n+1)n3=3n2+3n+1tn+1=tn+3n2+3n+1 which matches the original definition of the sequence. We have proved that the original sequence can be expressed as tn=n3, and thus, t100=1003.

  3. If a, b, a+b, and ab are positive numbers that form 4 consecutive terms in a geometric sequence, find a.
    Solution

    Since we have a geometric sequence, the ratios of consecutive terms will be equal. So ()ab=ba+b=a+bab Therefore, ab=ba+ba2+ab=b2b2aba2=0(ba)2(ba)1=0(since a0)ba=1+52 where we have chosen the positive root since a and b are positive. Also from (), ab=a+baba2=a+ba=1+ba(since a0)=3+52(substituting from above)

Problem Set

  1. In a geometric series, t5+t7=1500 and t11+t13=187500. Find all possible values for the first three terms.

  2. Given that a, b and c are consecutive terms in an arithmetic sequence that has distinct terms, calculate x if (bc)x2+(ca)x+(ab)=0

  3. If x, 4, y are consecutive terms in an arithmetic sequence and x, 3, y are consecutive terms in a geometric sequence, calculate 1x+1y.

  4. Three different numbers, whose product is 125, are 3 consecutive terms in a geometric sequence. At the same time they are the first, third and sixth terms of an arithmetic sequence. Find these three numbers.

  5. The kth triangular number is given by Tk=1+2+3++k=k(k+1)2=k2+k2. The first six triangular numbers are 1, 3, 6, 10, 15, and 21. Find the sum of the first 200 triangular numbers.

  6. If the interior angles of a pentagon form an arithmetic sequence and one interior angle is 90°, find all possible values of the largest angle in the pentagon.

  7. Find the four integers a, b, c and d that satisfy the following conditions:

  8. A sequence t1, t2, t3 is formed by choosing t1 at random from the set {1,2,3}, t2 at random from the set {4,5,6}, and t3 at random from the set {7,8,9}. What is the probability that t1, t2, t3 is an arithmetic sequence?

  9. The sum of 25 consecutive integers is 500. Determine the smallest of the 25 integers.

  10. What is the number of terms in the arithmetic sequence 1994,1992,1990,,1992,1994?

  11. The sum of the first n terms of a sequence is Sn=3n1, where n is a positive integer.

    1. If tn represents the nth term of the sequence, determine t1, t2, t3.

    2. Prove that tn+1tn is constant for all values of n.

  12. How many terms in the arithmetic sequence 7,14,21, are between 40 and 28001?

  13. If f is a function such that f(1)=2 and f(n+1)=3f(n)+13 for n=1, 2, 3,, what is the value of f(100)?

  14. Consider the family of lines with equations of the form px+qy=r, and which all pass through the point (1,2). Prove that p, q, and r are consecutive terms of an arithmetic sequence.

  15. An arithmetic sequence S has terms t1,t2,t3,, where t1=a and the common difference is d. The terms t5, t9, and t16 form a three-term geometric sequence with common ratio r. Prove that S contains an infinite number of three-term geometric sequences, all having the same common ratio r.

  16. In the sequence 5,3,2,5,, each term after the first two is constructed by taking the preceding term and subtracting the term before it. What is the sum of the first 32 terms in the sequence?

  17. Consider the sequence t1=1, t2=1 and tn=(n3n1)tn2 where n3. What is the value of t1998?

  18. The nth term of an arithmetic sequence is given by tn=5557n. If Sn=t1+t2++tn, determine the smallest value of n for which Sn<0.