Sequences and Series · Lesson 3

Sums of arithmetic and geometric series

A series is the sum of the terms of a sequence. For an arithmetic series, pair the first and last terms: each pair has the same total, so Sn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}. In words, the sum is the number of terms times the average of the first and last terms.

Before you use the formula, count the terms. The integers from 15 to 75 inclusive number 75−15+1=6175 - 15 + 1 = 61, not 60. For a list such as 3, 7, 11, ..., 83, find nn from an=a1+(n−1)da_n = a_1 + (n - 1)d.

For a geometric series, questions usually give the formula, Sn=a1(1−rn)1−rS_n = \frac{a_1(1 - r^n)}{1 - r} for r≠1r \neq 1, but know it in case one doesn't. Substitute carefully: nn is the number of terms, and the exponent on rr here is nn, not n−1n - 1.

Series formulas
SeriesSum of the first nn termsExample
ArithmeticSn=n(a1+an)2S_n = \frac{n(a_1 + a_n)}{2}1+3+⋯+99=50(1+99)2=2,5001 + 3 + \cdots + 99 = \frac{50(1 + 99)}{2} = 2{,}500
Geometric (r≠1r \neq 1)Sn=a1(1−rn)1−rS_n = \frac{a_1(1 - r^n)}{1 - r}2+4+8+16=2(1−24)1−2=302 + 4 + 8 + 16 = \frac{2(1 - 2^4)}{1 - 2} = 30

Worked example Easy

What is the sum of the first 50 positive odd integers?

  1. 9999
  2. 2,5002{,}500Answer
  3. 2,5502{,}550
  4. 5,0005{,}000

How to solve it

  1. The odd integers form an arithmetic sequence with a1=1a_1 = 1 and d=2d = 2.
  2. The 50th term is 1+49⋅2=991 + 49 \cdot 2 = 99.
  3. S50=50(1+99)2=25⋅100=2,500S_{50} = \frac{50(1 + 99)}{2} = 25 \cdot 100 = 2{,}500.
  4. Pattern check: the sum of the first nn odd integers is n2n^2, and 502=2,50050^2 = 2{,}500.

Why each choice is right or wrong

  • A. Incorrect. 99 is the 50th odd integer, the last term, not the sum of all 50 terms.
  • B. Correct. a50=99a_{50} = 99, so S50=50(1+99)2=2,500S_{50} = \frac{50(1 + 99)}{2} = 2{,}500.
  • C. Incorrect. 2,550 uses 101 as the last term, from 1+50⋅21 + 50 \cdot 2. The 50th term is 49 steps after the 1st, so it is 99.
  • D. Incorrect. 5,000 is 50(1+99)50(1 + 99) without dividing by 2. The formula multiplies nn by the average of the first and last terms.

Worked example Medium

Rosa saves 3 dollars in the first week and then saves twice as much each week as the week before. The total saved in the first nn weeks is Sn=a1(1−rn)1−rS_n = \frac{a_1(1 - r^n)}{1 - r}, where a1a_1 is the first week's amount and rr is the common ratio. How many dollars does she save in total over the first 8 weeks?

  1. 381381
  2. 384384
  3. 765765Answer
  4. 1,5331{,}533

How to solve it

  1. The amounts form a geometric sequence with a1=3a_1 = 3 and r=2r = 2, and n=8n = 8.
  2. S8=3(1−28)1−2=3(1−256)−1S_8 = \frac{3(1 - 2^8)}{1 - 2} = \frac{3(1 - 256)}{-1}.
  3. =3(−255)−1=765= \frac{3(-255)}{-1} = 765.

Why each choice is right or wrong

  • A. Incorrect. 381 uses 272^7 in the sum formula, which is the total for 7 weeks. In SnS_n the exponent is n=8n = 8.
  • B. Incorrect. 384 is 3⋅273 \cdot 2^7, the amount saved in week 8 alone. The question asks for the total of all 8 weeks.
  • C. Correct. S8=3(1−256)1−2=−765−1=765S_8 = \frac{3(1 - 256)}{1 - 2} = \frac{-765}{-1} = 765 dollars.
  • D. Incorrect. 1,533 uses 292^9, which is the total for 9 weeks. Count the weeks: n=8n = 8.

Trap Counting the terms

From 15 to 75 inclusive there are 75−15+1=6175 - 15 + 1 = 61 integers. Subtracting the endpoints gives the number of steps, which is one less than the number of terms. When a choice is off by exactly one term's worth, recheck your count.

Trap A term is not a sum

The last term ana_n and the sum SnS_n are different numbers. The amount saved in week 8 is 3⋅27=3843 \cdot 2^7 = 384, but the total over 8 weeks is 765. Read whether the question asks for one term or the total.

Calculator When a calculator is faster

Once you have nn, a1a_1 and ana_n, a calculator does the arithmetic quickly, and it handles a power such as 282^8 inside the geometric formula. In the built-in Desmos calculator, if you test online, you can also check a sum directly by typing it with the sum template. Counting the terms is still something you do yourself.