Sequences and Series · Lesson 1

Arithmetic sequences

An arithmetic sequence adds the same number, the common difference dd, to get from each term to the next. In 5, 11, 17, 23, ... the difference is d=6d = 6. A negative dd makes the terms fall: 40, 37, 34, ... has d=−3d = -3.

From the 1st term to the nnth term takes n−1n - 1 steps, so an=a1+(n−1)da_n = a_1 + (n - 1)d. The 30th term is 29 steps after the 1st, not 30. This one idea is behind most wrong answers in this topic.

When you know two terms that are not next to each other, count the steps between them: from the 4th term to the 10th term is 10−4=610 - 4 = 6 steps. Divide the change in value by the number of steps to get dd, then step backward or forward to the term you need.

Worked example Easy

An arithmetic sequence begins 5, 11, 17, 23, ... . What is the 30th term of the sequence?

  1. 174174
  2. 179179Answer
  3. 180180
  4. 185185

How to solve it

  1. The common difference is 11−5=611 - 5 = 6.
  2. The 30th term is 30−1=2930 - 1 = 29 steps after the 1st: 29⋅6=17429 \cdot 6 = 174.
  3. Add the first term: a30=5+174=179a_{30} = 5 + 174 = 179.

Why each choice is right or wrong

  • A. Incorrect. 174 is the total of the 29 steps, 29⋅629 \cdot 6. You still need to add the first term, 5.
  • B. Correct. a30=5+(30−1)⋅6=5+174=179a_{30} = 5 + (30 - 1) \cdot 6 = 5 + 174 = 179.
  • C. Incorrect. 180 is 30⋅630 \cdot 6: it counts one step too many and leaves out the first term.
  • D. Incorrect. 185 is 5+30⋅65 + 30 \cdot 6, which uses 30 steps. Only 29 steps separate the 1st term from the 30th.

Worked example Medium

In an arithmetic sequence, the 4th term is 19 and the 10th term is 49. What is the 25th term of the sequence?

  1. 124124Answer
  2. 129129
  3. 139139
  4. 145145

How to solve it

  1. From the 4th term to the 10th term is 10−4=610 - 4 = 6 steps, and the value rises 49−19=3049 - 19 = 30.
  2. So d=306=5d = \frac{30}{6} = 5.
  3. Step back to the 1st term: a1=19−3⋅5=4a_1 = 19 - 3 \cdot 5 = 4.
  4. Then a25=4+24⋅5=124a_{25} = 4 + 24 \cdot 5 = 124. Faster: a25=a10+15⋅5=49+75=124a_{25} = a_{10} + 15 \cdot 5 = 49 + 75 = 124.

Why each choice is right or wrong

  • A. Correct. d=49−1910−4=5d = \frac{49 - 19}{10 - 4} = 5 and a1=4a_1 = 4, so a25=4+24⋅5=124a_{25} = 4 + 24 \cdot 5 = 124.
  • B. Incorrect. 129 is 4+25⋅54 + 25 \cdot 5, which takes 25 steps from the 1st term. The 25th term is 24 steps after the 1st.
  • C. Incorrect. 139 is 19+24⋅519 + 24 \cdot 5, which starts from the 4th term as if it were the 1st. From the 4th term to the 25th is only 21 steps.
  • D. Incorrect. 145 divides the change by 5, the number of terms strictly between the 4th and 10th, giving d=6d = 6. The gap is 6 steps, so d=5d = 5.

Trap Counting terms instead of steps

The nnth term is n−1n - 1 steps after the first. Choices built from a1+nda_1 + nd, or from n⋅dn \cdot d with no first term, are waiting for students who count terms. Test your rule on a term you can see: for 5, 11, 17, the 3rd term should be 5+2⋅6=175 + 2 \cdot 6 = 17.

Trap Dividing by the wrong gap

From the 6th term to the 15th term is 15−6=915 - 6 = 9 steps. Counting the terms from 6 to 15 gives 10, and counting only the terms strictly between them gives 8; neither is the number of steps. Subtract the positions, and divide the change in value by that.

Calculator When a calculator isn’t faster

These items are one or two lines of arithmetic, so hand work is usually quickest. Your calculator helps with a product such as 29⋅629 \cdot 6, and in the built-in Desmos calculator, if you test online, you can type a rule such as 5+(n−1)⋅65 + (n - 1) \cdot 6 into a table to list terms and check your count.