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David Terr
Ph.D. Math, UC Berkeley

 

Home >> Pre-Calculus >> 10. Sequences, Induction, and Probability

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10.2. Arithmetic Sequences

In the previous section, we introduced the reader to sequences and series. In this section, we will study a special type of sequnces known as arithmetic sequences, as well as their associated series.

An arithmetic sequence is a sequence of the form a0 = a, a1 = a + b, a2 = a + 2b, a3 = a + 3b, ..., where a and b are constants. Clearly, the term of the sequence with index k is ak = a + kb. The defining property of an arithmetic sequence is that the difference between consecutive terms is constant. The whole numbers, counting numbers, even numbers, odd numbers, and multiples of 3 are all examples of arithmetic sequences.

It is easy to compute the term with index k of an arithmetic sequence. We know it is equal to a + kb, so all we have to do is to determine a and b. But a is just the first term of the sequence and b is the difference between consecutive terms, positive if the sequence is increasing and negative if the sequence is decreasing.

 

Example 1: For each of the following arithmetic sequences, compute a10 and a20. (Assume each sequence starts with index 0.)

  • (a) 1, 3, 5, 7, ...
  • (b) 0, 5, 10, 15, 20, ...
  • (c) 5, 10, 15, 20, ...
  • (d) 1, 4, 7, 10, ...
  • (e) 100, 99, 98, 97, ...

Solution:

  • (a) The sequence begins with a = 1 and the difference between consecutive terms is b = 2. Therefore, we have ak = a + kb = 1 + 2k. Thus, we have a10 = 1 + (2)(10) = 21 and a20 = 1 + (2)(20) = 41.
  • (b) The sequence begins with a = 0 and the difference between consecutive terms is b = 5. Therefore, we have ak = a + kb = 5k. Thus, we have a10 = (5)(10) = 50 and a20 = (5)(20) = 100.
  • (c) The sequence begins with a = 5 and the difference between consecutive terms is b = 5. Therefore, we have ak = a + kb = 5 + 5k. Thus, we have a10 = 5 + (5)(10) = 55 and a20 = 5 + (5)(20) = 105.
  • (d) The sequence begins with a = 1 and the difference between consecutive terms is b = 3. Therefore, we have ak = a + kb = 1 + 3k. Thus, we have a10 = 1 + (3)(10) = 31 and a20 = 1 + (3)(20) = 61.
  • (e) The sequence begins with a = 100 and the difference between consecutive terms is b = -1. Therefore, we have ak = a + kb = 100 - k. Thus, we have a10 = 100 - 10 = 90 and a20 = 100 - 20 = 80.

 

There is a simple formula for computing series corresponding to arithmetic sequences, known as arithmetic series.. Let a0 = a, a1 = a + b, a2 = a + 2b, a3 = a + 3b, ... be an arithmetic sequence. Then we have

  • (10.2.1) arithmetic series formula

It is not too difficult to derive this formula. The trick is to match up pairs of terms, starting with the beginning and the end of the sequence. This sum is equal to {a + (a + nb)} + {(a + b) + [a + (n-1)b]} + {(a + 2b) + [a + (n-2)b]} + ... . Note that each term in braces is equal to 2a + nb. Now if n is odd, then all terms in the sequence are paired in this way into (n+1)/2 pairs, implying that the sum of the sequence is (n+1)(2a+nb)/2 as claimed. What if n is even? Then there are n/2 pairs and one term left over in the middle of the sequence, equal to a + nb/2. Thus, in this case, the sum is equal to n/2(2a + nb) + a + nb/2 = (n/2 + 1/2)(2a + nb) = (n+1)(2a+nb)/2. This proves the formula.

As a special case of Equation (10.2.1), we derive a formula for the sum of the first n counting numbers. For such a sequence, we have a = b = 1 and we must replace n with n-1 since n is one less than the number of terms. The result thus becomes n[(2)(1) + (n-1)(1)] / 2 = n(2 + n - 1) / 2 = n(n+1)/2. Thus we have

  • (10.2.2) natural number series

 

Example 2: Use Equation (10.2.2) to compute the sum of the first 100 counting numbers.

Solution: Plugging in n=100 into Equation 10.2.2, we find the sum is equal to (100)(101)/2 = 5050.

 

Example 3: Use Equation (10.2.1) to compute the sum of the first 100 terms of each arithmetic sequence from Example 1. (This means plugging in n=99.)

Solution:

  • (a) The sum is equal to (99+1)[(2)(1) + (99)(2)] / 2 = (100)(2+198)/2 = (100)(200)/2 = 10,000.
  • (b) The sum is equal to (99+1)[(2)(0) + (99)(5)] / 2 = (100)(0+495)/2 = (100)(495)/2 = 24,750.
  • (c) The sum is equal to (99+1)[(2)(5) + (99)(5)] / 2 = (100)(10+495)/2 = (100)(505)/2 = 25,250.
  • (d) The sum is equal to (99+1)[(2)(1) + (99)(3)] / 2 = (100)(2+297)/2 = (100)(299)/2 = 14,950.
  • (e) The sum is equal to (99+1)[(2)(100) + (99)(-1)] / 2 = (100)(200-99)/2 = (100)(101)/2 = 5050.

 

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