How to Crack Mathematical Sequences & Patterns

In SAT Math by Peter PengLeave a Comment

Humans are pattern-recognition machines. That’s why the SAT likes to test how good you are at seeing them.

There are four general types of sequence/pattern questions on the SAT.

  1. Arithmetic Sequences
  2. Geometric Sequences
  3. Repeating Sequences
  4. Miscellaneous Patterns

The first two (arithmetic and geometric) have formulas you MUST memorize. The latter two (repeating and miscellaneous) have special approaches you should use but no formulas.

Arithmetic Sequences: a pattern involving adding or subtracting the same number repeatedly.

E.g. 2, 4, 6, 8, 10…(+2 between every term)

E.g. 15, 10, 5, 0, -5, -10… (-5 between every term)

Formula: An = A1 + d (n-1) <– IMPORTANT: it’s always a + (plus sign) after A1, never a – (minus sign)

An = the value of the nth term

A1 = the value of the 1st term

n = the term you’re trying to calculate (so if n is 5, then A5 means the value of the fifth term in the sequence)

d = the distance between each term (so if you’re adding the same number each time, then d is positive…if you are subtracting, then d is negative)

Basically, if you think about the formula in English, it goes like this:

Value of the nth term = value of 1st term + (distance between each term × how many times you need to apply that distance)

Check it out in this example:

Given the following sequence 3, 6, 9, 12, etc. what is the 100th term?

A100 = A1 + 3 (100 – 1)

The formula works this way because every sequence starts with the first term (A1, which is 3 in this case), then you keep adding the same number, d, to A1 over and over. For example, A1 + 3 + 3 + 3 + 3…

But rather than writing +3 out so many times, notice that 3 + 3 + 3 + 3 = 3 x 4. Remember that? What we just did is apply the definition of multiplication you learned in second grade.

So what we need to figure out is how many times we need to add 3 to our first term (A1). Notice the d (n-1) part of the formula. d is how much—NOT how many times—you add each time (in our case, you add 3 each time). The (n-1) represents how MANY times you need to add 3. So if we’re looking for the 100th term, we will add 3 ninety-nine times because 100 – 1 = 99. Why do we always subtract 1? Because you don’t add anything to get to the first term (A1). The first term ALREADY exists on its own. The very first 3 you add will already get you to the second term. If you add 3 twenty times, that will get you the 21st term. If you add 3 forty-five times, that will get you the 46th term. So if you add 3 ninety-nine times, you will get the 100th term. Capache?

An = A1 + d (n-1)

A100 = A1 + 3 (100-1)
A100 = 3 + 3 (99)
A100 = 3 + 297
A100 = 300.

The 100th term is 300. THIS IS SPARTAAAA!!!

SUMMATION TRICK: There’s another devilish thing the SAT sometimes loves to pull. Occasionally, the SAT wants you to calculate the SUM of a bunch of terms in an arithmetic sequence.

For example, let’s say we have this sequence: 1, 3, 5, 7….99. They want you to add everything in that sequence up. Are you seriously going to do 1 + 3 + 5 + 7…+ 99? Do you have special powers that allow you to freeze time or something? No. Didn’t think so, so listen, we don’t have all day to do that tedious manual addition method. It’ll work, yes, but I’m lazy and don’t have time for such shenanigans. Pro tip: use every shortcut you can find on the SAT to save time.

So what are we going to do? Hmm. Well, notice this:

1 + 3 is the same as 3 + 1, right? So it doesn’t matter what order I add all the terms up. I just have to eventually add all of them up. What if I did 1 + 99 first? That gets me 100. Now what if I went one term in from the left and one term in from the right? That gets me 3 + 97, which is still 100. Now let’s go two terms in from the left and two from the right…5 + 95 = 100. See it yet? It’s always 100, and therein lies the magic, my friend.

If you keep pairing terms up starting with the outermost terms and slowly making your way towards the center, you will always have a sum of 100 per pair. It’s like a school dance—everyone wants a partner to dance with, so go find your crush and dance with him or her. Life tip: smell nice and brush your teeth because you’re going to be leaning in close to each other.

1 + 3 + 5 + 7 …. + 93 + 95 + 97 + 99 = ?

Well, let’s do:

1 + 99
3 + 97
5 + 95…and so on.

Each pair of terms is worth a sum of 100. But how many pairs do we have altogether? Well, let’s figure out how many terms total there are, then divide by two (to pair things up).

PRO TIP: The best way to count the number of terms in a longer sequence is to examine a shorter version with the same pattern, then extrapolate your findings back to the longer sequence.

Let’s look at the short sequence 1, 3, 5, 7. It has 4 terms (2 pairs).

Another short sequence with the same pattern: 1, 3, 5, 7, 9. It has 5 terms (2.5 pairs).

Notice if I add the outermost numbers together, then divide by 2, that’s how many terms in total I have.

In the first short sequence, 1 + 7 = 8. Divide that by 2 = 4 terms total.
In the second short sequence, 1 + 9 = 10. Divide that by 2 = 5 terms total.

I can take this idea and use it on the longer sequence: 1, 3, 5, 7…95, 97, 99.

So 1 + 99 = 100…divide by 2 = 50 terms total. Remember, our goal is to calculate the sum of all 50 of these terms. If we have 50 terms, then there are half as many pairs: 25.

25 pairs x 100 (which is the sum of each pair) = the sum of all pairs added together = your nifty answer.

25 x 100 = 2,500. BOOM!

ONE MORE THING…DON’T FORGET THE LONELY GUY

Well, the previous question was nice and easy because there was an even number of terms, hence an even number of pairs. But what if there were an odd number of terms? Then not everyone would get to pair up right? The creepy guy in the middle of the sequence would be left all alone…no one wants to dance with him because he didn’t take my advice about brushing his teeth and smelling awesome. Yet, we still gotta play nice and add him too.

Imagine this: 1 + 3 + 5….+ 101 = ?

We actually went over 100 this time and still skip every other integer. That means there are 51 terms. An easy way to calculate the total number of terms in a sequence here is to add the first and last term, then divide by 2 (because we want to toss out all the values we are skipping over). 1 + 101 = 102. Then 102/2 = 51 total terms.

Since 51 is an odd number of terms, there’s going to be a lonely dude in the middle. Who is he? Well, he’s the average of the two outermost terms, so the lonely term is going to be 51. That lonely term is calculated like this: (1 + 101)/2 = 51. Don’t forget to add 51 at the end of the problem.

Now we do the same as before. We rearrange the order in which we add the terms, combining as many as we can into pairs.

1 + 101 = 102
3 + 99 = 102
5 + 97 = 102
…and so on.

With 51 total terms, we can only get 25 full pairs plus the lonely leftover guy (51).

So 25 full pairs x 102 (which is the sum of each pair) + 51 (aka the lonely guy with no dance partner) = total sum of the whole sequence.

25×102 + 51 = 178. That’s it! So remember, play nice and don’t forget to include the lonely guy.

 

 

Geometric Sequences: a pattern involving multiplying or dividing by the same number repeatedly.

E.g. 2, 4, 8, 16, 32…(×2 between every term)

E.g. 81, 27, 9, 3, 1, 1/3, 1/9… (÷ 3 between every term)

Formula: An = A1 ×  r^(n-1) IMPORTANT: it’s always a × (multiplication sign) after A1, never a ÷ (division sign)

An = the value of the nth term

A1 = the value of the 1st term

n = the term you’re trying to calculate (so if n is 5, then A5 means the value of the fifth term in the sequence)

r = the ratio between each term (if you’re multiplying by 3 each time, r is 3…if you are dividing each time, then r is)

Basically, if you think about the formula in English, it goes like this:

Value of the nth term = value of 1st term × (ratio between each term raised to the power of how many times you need to apply that ratio)

How does the formula work? Well, we always start with the first term, A1, duh. Then from there, we keep multiplying A1 by a number, which is represented by r. So let’s say the first term in our sequence is 2 and the pattern is we keep multiplying by 3. So basically what we’re doing is 2 × 3 × 3 × 3 × 3…a bunch of times until we get to the term we are trying to calculate.

Remember how exponents work? 3 × 3 × 3 × 3 = 3^4

So 2 × 3 × 3 × 3 × 3 really is the same thing as 2 × 3^4.

But why do we subtract 1 in the exponent power part (n-1)? Because the first term already exists on its on BEFORE you ever multiplied it by 3. The first term is really itself multiplied by 3^0 (remember, anything raised to the 0 power just equals 1; so A1 × 1 still equals A1 itself). The first time you multiply A1 by 3, you are going to get the 2nd term. The second time you multiply by 3, you are going to get the 3rd term. Notice how whatever term you’re looking for is always 1 less than the number of times you multiplied by 3.

To calculate, say, the 15th term, you only need to multiply A1 by 3 fourteen times. Since A1 itself is already the first term, so you don’t need to raise the 3 by an extra power. By subtracting 1 from n, you are ensuring that you’re not erroneously taking r to an extra power.

Let’s try an example.

Example Question: Find the 20th term of the following sequence: 81, 27, 9, 3, 1, 1/3, 1/9…

First realize this is a geometric, not an arithmetic, sequence because it’s dividing by 3 between every term. That means our r = 1/3 because dividing by 3 is the same thing as multiplying by 1/3. So use the geometric sequence formula:

An = A1 × r ^ (n-1)

A20 = A1 ×  (1/3) ^ (n – 1)

A20 = 81 ×  (1/3) ^ (20 – 1)

A20 = 81 ×  (1/3) ^ 19

A20 = 81 × (1/1162261467) <– answer

You can simplify this as much as you want, but it’s a pretty small number. The SAT will likely stop at  A20 = 81 × (1/3) ^ 19.

 

 

Repeating Sequences: a pattern that is neither arithmetic or geometric, but still possess a pattern. The pattern repeats the same set of numbers or the same set of number properties over and over.

Keep in mind, there are NO FORMULAS for repeating sequences. You just have to learn the special approach to deal with repeating sequences.

Imagine this sequence: 1, 1, 0, 1, 1, 0, 1, 1, 0…

It’s not adding or subtracting the same number each time, so it’s not an arithmetic sequence. It’s also not multiplying or dividing by the same number each time, so it’s not a geometric sequence. But it is repeating 1, 1, 0 over and over. Thus, behold the marvelous repeating sequence. I say that as if it’s magical, but really, I just happen to think this is one of the coolest math concepts on the SAT. Yeah yeah, I’m a nerd. Eat it.

What if they want us to find out the 1,000th term of this repeating sequence? It’s clear it’s got to be either 1 or 0 because those are the only two terms this sequence repeats, but we’re not going to write out the sequence all the way to 1,000 terms. That’s insane and inane.

Now what? Well, follow along this 5-step process:

STEP 1: circle the repeating block

STEP 2: count how many terms there are in one block (every block will have the same number of terms)

STEP 3: divide whatever term you want to calculate by the number of terms in each block (aka what you found in STEP 2)

STEP 4: ignore the decimal portion of STEP 3’s result, if any. But take the whole number portion you found in STEP 3 and multiply it by the number of terms in each block (STEP 2’s result). This result represents the total number of terms in all the full, complete blocks of repeating terms that can go into whatever ultimate term you’re trying to calculate.

STEP 5: manually count over the next few terms until you reach the ultimate term you desire. That’s your answer!

 

If that above was a little hard to follow, then check out the example with actual numbers:

STEP 1: circle 1, 1, 0 because that’s the repeating block of terms

STEP 2: count that there are 3 terms per repeating block

STEP 3: I’m looking for the 1,000th term, so I’m going to divide 1,000 terms by 3 terms per block. That gets me 333.333333.

STEP 4: Ignore the .33333 part to the right of the decimal. Just focus on the whole number part (333) to the left of the decimal. That means there are 333 full, complete repeating blocks that can go into 1,000 terms without exceeding it. So take 333 full blocks and multiply by 3 terms per block = 999 total terms in all the full blocks.

STEP 5: I’m looking for the 1,000th term, but STEP 4 only told me the 999th term was the last term to complete a full block. Because the last term of any and all full blocks is 0, I know the 999th term is 0. Now, the 1,000th term is just one term after the 999th term, so just count one term over…which becomes 1, the start of a NEW block. The value of the 1,000th term is 1.

There you have it!

 

ANOTHER LAYER OF HELL

Now, if you truly want to master this repeating sequences concept, you can’t stop yet. The SAT has another level of hell waiting for you. You haven’t faced the final boss yet. This is the complication they might throw at you:

Let’s say we have the same pattern: 1, 1, 0, 1, 1, 0, etc. But now they want to know how many 1s there are in the first 1,000 terms.

Earlier, we found out the 1,000th term is 1. Well, to find out how many instances of 1 show up in the first 1,000 terms, we again have to figure out how many full, complete blocks go into 1,000 terms. We calculated that in STEP 4. We have 333 full, complete blocks that comprise 999 individual terms. Notice how we scoot right up to 1,000 terms as close as possible, but we never exceed it. If we exceeded 1,000, then we are calculating with too many terms now.

Okay, so think about how many instances of 1 there are per full block. Two, because there are two 1s in the block of 1, 1, 0. That means for every full block, there are two 1s. Follow me here? If not, go re-read what I just wrote.

So each full block has two 1s, and we have 333 full blocks. That means 2 x 333 = number of 1s in 333 full blocks. There are 666 instances of 1 in 333 full blocks, which is the same as saying there are 666 instances of 1 in the first 999 terms. Now look at the 1,000th term, which is also 1. Since the question asks how many 1s there are in the first 1,000th term, we must include that last instance of 1.

666 + 1 = 667 instances of 1 in the first 1,000 terms. That’s our answer.

IMPORTANT NOTE: If the 1,000th term somehow weren’t 1, then we wouldn’t include another instance of 1. The answer would only be 666. But since the 1,000th term is 1, the real answer is 667.

 

WHEN YOU DON’T INCLUDE THE LAST TERM(S)
Here’s an example of when you don’t include the last term. Imagine they want us to calculate how many 0s there are in the first 1,000 terms now.

1, 1, 0, 1, 1, 0…etc.

The repeating block is 1, 1, 0, which has 3 terms per block. There is one instance of 0 per full, complete block. We know the 1,000th term is 1 from our earlier calculations. We also know from earlier that there are 333 full, complete blocks that go into 1,000 total terms. These 333 full blocks contain 999 total terms. That means there is 1 instance of 0 per block x 333 blocks = 333 instances of 0 in all the full blocks altogether. And since the 1,000th term is not 0, we don’t add another term. Our answer is simply 333 instances of 0 in the first 1,000 terms. The number of 0s in the first 1,000 terms is the SAME as the number of 0s in the first 999 terms because the 1,000 term doesn’t count (it’s not a 0).

 

REPEATING PROPERTIES: INCEPTION DREAM LAYERS

So that last question was pretty difficult, but we destroyed that monster. Unfortunately, the SAT is like a hydra…you cut one head off, two more grow in its place. There’s actually one more level to this thing, the hardest and final level for these types of questions, I promise. They are the REPEATING PROPERTIES PATTERNS.

These are sequences or patterns that don’t repeat the same numbers, but repeat the same properties of those numbers. Ever see the movie Inception with Leonardo DiCaprio where they keep going down into deeper levels of their dreams? It’s kind of like that.

Look at this sequence: 5, 3, 0, 117, -1, 22, 5, 7, -22…

At first glance, there is no pattern. It’s not arithmetic, geometric, or repeating, right? Well…actually, it is repeating something, just not the actual numbers themselves.

5, 3, 0, 117, -1, 22, 5, 7, -22…

ODD, ODD, EVEN, ODD, ODD, EVEN, ODD, ODD, EVEN…see it now?

The PROPERTIES of these numbers are either even or odd. The repeating block is “odd, odd, even.”

How did I know to even think of odd and even? Because the question will ALWAYS tell you want to look for. They will ask how many odd terms are there in the first 1,000 terms. You approach this the same way as you do the normal repeating patterns, but instead of caring about the value of the actual numbers, let’s just convert everything to the deeper level of odds and evens.

You can rewrite the sequence if it helps: O, O, E, O, O, E, etc. We really don’t give two craps about what the value of the actual numbers are because the question is not asking about the numbers themselves, but the even or odd PROPERTIES of the numbers.

I think you can handle the calculations from here, yes? Go through the same 5-step process as for normal repeating sequences and go through the “Another Layer of Hell” section.

2 odds per block x 333 full blocks = 666 odds in the first 999 terms.

Since the 1,000th term is also odd, we have to include another term. So 666 + 1 = 667 odd numbers in the first 1,000 terms.

 

Miscellaneous Patterns: these are just patterns that are not any of the types we’ve talked about above.

Example question:
1.101001000100001000001…

In the number above, the first 1 to the right of the decimal is followed by one 0, the second 1 to the right of the decimal is followed by two 0s, the third 1 is followed by three 0s, and so on. How many 0s are there between the 98th and 101st 1 to the right of the decimal?

Clearly, there’s a defined pattern here, but it’s nothing like the other patterns we’ve seen thus far. Not to panic. Just break down the question phrase by phrase.

They keep mentioning the right of the decimal, so we really don’t care about the 1 to the left of the decimal. So pretend the 1 to the left has disappeared, leaving .101001000100001000001…

After the first 1, there is one 0.
After the second 1, there are two 0s.
After the third 1, there are three 0s.

Easy peasy. The question is asking how many 0s there are BETWEEN the 98th 1 and the 101st 1.

So after the 98th 1, there are ninety-eight 0s.
After the 99th 1, there are ninety-nine 0s.
After the 100th 1, there are a hundred 0s.
After the 101th 1, well…we don’t care because the question is asking how many 0s are BETWEEN the 98th and 101st 1. All the 0s after the 101st 1 are not between the specified range, so ignore them!

So, 98 + 99 +100 = 297 zeroes! Nailed it!

Well, so long for now. “Live long and prosper” as the great Spock from Star Trek once said. You’ll live to fight another battle on this grueling journey that is the SAT.

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