standard_description: Compare two fractions with different numerators and different denominators, e.g., by creating common denominators or numerators, or by comparing to a benchmark fraction such as 1/2. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with symbols >, =, or <, and justify the conclusions, e.g., by using a visual fraction model.
Grade 4 expectations in this domain are limited to fractions with denominators 2, 3, 4, 5, 6, 8, 10, 12, and 100.
problem_activity: Warm-up
This warm-up picks up where the previous lesson ended asking students to calculate the sum of the first 50 terms of two different sequences, but with less scaffolded support. The goal of this activity is to build student flexibility in using the formula for the sum and thinking about the terms in a sequence by comparing two related sequences, specifically, two sequences where one has values double that of the first.
Monitor for students who identify the sum of the second sequence in different ways, such as by
thinking of it as a sequence with the same
\(r\)
and
\(n\)
values as the first but a different
\(a\)
value.
noticing it must be twice the sum of the first.
seeing it as 1 plus the sum of the first sequence, only where the first sequence has
\(n=49\)
instead of
\(n=50\)
.
Student Facing
Recall that for any geometric sequence starting at
\(a\)
with a common ratio
\(r\)
, the sum
\(s\)
of the first
\(n\)
terms is given by
\(s=a\frac{1-r^{n}}{1-r}\)
. Find the approximate sum of the first 50 terms of each sequence:
\(\frac12\)
,
\(\frac14\)
,
\(\frac18\)
,
\(\frac{1}{16}\)
, . . .
1,
\(\frac12\)
,
\(\frac14\)
,
\(\frac18\)
,
\(\frac{1}{16}\)
, . . .
Student Response
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Activity Synthesis
Once students are agreed on the
\(a\)
,
\(r\)
, and
\(n\)
values for the first sequence, invite previously selected students to share their work for the second sequence. While students describe their thinking, record any expressions used for all to see. If not brought up by a student for the second strategy, display the relationship
\(1 \frac{1-\left(\frac12\right)^{n}}{1-\frac12} = 2 \boldcdot \frac12 \frac{1-\left(\frac12\right)^{n}}{1-\frac12}\)
as one way to understand why the second sequence has a sum twice as large.


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