So far, we have discussed solving for X in 5 different scenarios: SolveX, SolveXExpo, SolveXDenom, SolveXFacto, and SolveX2var. Today we will look at sequences with unknowns. On middle school tests, this has appeared 12 times so far this year, with a median placement at question # 44.
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For the most part, these problems need to be approached like any other sequence problem. It might be a good idea to review these sequence concepts before proceeding:
- AddSequence (the "catch-all" formula for arithmetic sequences)
- AddSeqEven (for consecutive even integers)
- AddSeqOdd (for consecutive odd integers)
In most cases, instead of going through every step of adding the sequence, you can save time by evaluating what the question is asking for. Let's look at some examples:
Example 1: If 2 + 4 + 6 + ... + 34 = 17k, then k = ___
- This is an sequence of even integers. To find the sum, we divide the last term by 2 and multiply it by its neighbor.
- 34 ÷ 2 = 17. We need to multiply 17 by 18 to get the sum (but don't take this step!).
- Note that the sum of the sequence is 17k. That means k = 18.
Example 2: If 3 + 6 + 9 + ... + 30 = 10k, then k = ___
- The sum of any arithmetic sequence is the number of terms times the median.
- There are 10 terms, and the median is (3 + 30)/2 = 33/2.
- If the sum is 10k, then k must be 33/2. (If the test doesn't specify what form to write the answer, leave it as an improper fraction instead of doing another calculation to write it as a mixed number or decimal).
Example 3: If 4 + 8 + 12 + ... + 48 = 12k, then k = ___
- (This problem was on the State test this year).
- There are 12 terms, and the median is (4 + 48)/2 = 52/2 = 26.
- The sum is 12k, so k must be 26.
Example 4: If 1 + 2 + 3 + ... + 41 = 21k, then k = ___
- There are 41 terms, and the median is (1 + 41)/2 = 42/2 = 21.
- The sum is 21k, so k must be 41.