SAT Sequences and Series

To solve problems about sequences and series on the SAT, you need to know how to

• find the \(n^\text{th}\) term in an arithmetic sequence: \( a_n = a_1 + (n-1)d \)
• find the \(n^\text{th}\) term in a geometric sequence: \(a_n = a_1 \times r^{n-1}\)
• find the sum of \(n\) consecutive terms in an arithmetic sequence: \[S_n = \frac{n(a_1+a_n)}{2}\]
• find the sum of \(n\) consecutive terms in a geometric sequence: \[S_{n}= a_1 \times \frac{r^n -1} {r-1}, \text{where}\ r\neq 1\]
• find a specific term of a sequence that is neither arithmetic nor geometric but follows a given pattern or rule
• find the number of terms in a sequence of consecutive integers.

If \(a\) and \(b\) are integers, and \(a, a+1, a+2,\ldots, b-1, b\) is a sequence of consecutive integers, then the number of terms between \(a\) and \(b\) is \(b-a+1\).

\[1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\]

In the sequence above, the first term is \(1\), and each successive term is obtained by multiplying the previous term by \(\frac{1}{2}\). What is the sum of the first \(8\) terms?

(A) \(\ \ 0\)

(B) \(\ \ \frac{1}{128}\)

(C) \(\ \ \frac{127}{64}\)

(D) \(\ \ \frac{255}{128}\)

(E) \(\ \ \frac{511}{256}\)

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Correct Answer: D

Solution 1:

Tip: Use a calculator.

\[1 + \frac{1}{2} +\frac{1}{4}+\frac{1}{8}+\frac{1}{16} +\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\frac{255}{128}.\]

Solution 2:

The sequence \(1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\) is geometric because the ratio between consecutive terms is the same, \(r=\frac{1}{2}\). Here we use the formula for finding the sum of \(n\) consecutive term of a geometric sequence. The first term is \(a_1=1\), and the number of terms is \(n=8\). Therefore,

\[\begin{array}{l c l l l} S_{8}&=& a_1 \times \frac{r^n-1} {r-1}\\ &=&1 \times \frac{\left(\frac{1}{2}\right)^{8}-1}{\frac{1}{2}-1} &\quad \text{apply formula} &(1)\\ &=&\frac{-\frac{255}{256}}{-\frac{1}{2}} &\quad \text{simplify} &(2)\\ &=&\left(-\frac{255}{256}\right)\times\left(-\frac{2}{1}\right) &\quad \text{multiply numerator}\\ &&&\quad \text{by reciprocal of denominator} &(3)\\ &=&\frac{255}{128}. &\quad \text{simplify} &(4)\\ \end{array}\]

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Incorrect Choices:

(A)
Tip: The simplest choice may not be the correct one.
Although the terms in the sequence decrease, they will always be positive because the product of two positive numbers is positive. Therefore, the sum of the first \(8\) terms will be greater than \(0\).

(B)
Tip: Read the entire question carefully.
If you solve for the \(8^\text{th}\) term in the sequence, instead of for the sum of the first \(8\) terms, you will get this wrong answer.

(C)
If you solve for the sum of the first \(7\) terms, instead of the sum of the first \(8\) terms in the geometric sequence, you will get this wrong answer.

(E)
If you solve for the sum of the first \(9\) terms, instead of the sum of the first \(8\) terms in the geometric sequence, you will get this wrong answer.

What is the sum of the consecutive integers from \(-49\) to \(94\)?

(A) \(\ \ 0\)
(B) \(\ \ 45\)
(C) \(\ \ 3217.5\)
(D) \(\ \ 3240\)
(E) \(\ \ 6480\)

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Correct Answer: D

Solution 1:

Tip: Use a calculator.
This is a time-consuming way to solve the problem.

\[(-49)+(-48)+(-47)+\cdots+91+92+93+94=3240.\]

Solution 2:

Tip: Look for shortcuts.
\((-49)+(-48)+(-47)+\cdots+91+92+93+94\) is an arithmetic sequence because the difference between two consecutive terms is constant. Let \(S\) be the sum of these consecutive integers. The number of terms in the sequence is \(n=94-(-49)+1=144\), the first term is \(a_{1}=-49\), and the last term is \(a_{n}=94\). Therefore,

\[S_{n}=\frac{n(a_{1}+a_{n})}{2}=\frac{144(-49+94)}{2}=\frac{144\cdot45}{2}=72\cdot45=3240.\]

Solution 3:

Tip: Look for shortcuts.
Let's call the unknown sum \(S\). Then

\[\small{S=\underbrace{(-49)+(-48)+\cdots+(-2)+(-1)+0+1+2+\cdots+48+49}_{=0}+50+\cdots+94}.\]

It should be immediately obvious that the sum of the integers from \(-49\) to \(49\) equals \(0\). If it isn't, note that we can pair opposite numbers together, and that opposite numbers add to \(0\), like this:

\[\small{S=\underbrace{(-49)+49}_{=0}+\underbrace{(-48)+48}_{=0}+\cdots+\underbrace{(-2)+2}_{0}+ \underbrace{(-1)+1}_{=0}+0+50+51+\cdots+93+94}.\]

\(S\) simplifies to \(S=50+51+52+\cdots+91+92+93+94\) and since it is an arithmetic series we can find its sum by using the formula \(S_{n}=\frac{n(a_{1}+a_{n})}{2}\).

In our case, the number of terms in the series is \(n=94-50+1=45\), the first term is \(a_{1}=50\), and the last term is \(a_{n}=94\). Therefore,

\[S_{n}=\frac{n(a_{1}+a_{n})}{2}=\frac{45(50+94)}{2}=\frac{45\cdot144}{2}=45\cdot72=3240.\]

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Incorrect Choices:

(A)
Tip: The simplest choice may not be the correct one.
This is the sum of the consecutive integers from \(-49\) to \(49\), but the question asks for the sum of the consecutive integers from \(-49\) to \(94\).

(B)
Tip: Read the entire question carefully.
If you solve for the sum of the two numbers in the prompt, \(-49\) and \(94\), instead of the sum of the consecutive integers from \(-49\) to \(94\), you will get this wrong answer.

(C)
Tip: Eliminate obviously wrong answers.
The sum of integers must be an integer, but this answer choice is not. You may get this wrong answer if when using the formula for finding the sum of \(n\) terms in an arithmetic sequence, you compute the number of terms from \(-49\) to \(94\) to be \(94-(-49)=143\). But the number of terms between \(-49\) to \(94\) is \(94-(-49)+1=144\).

(E)
If, when using the formula \(S_{n}=\frac{n(a_{1}+a_{n})}{2}\), you forget to divide by \(2\), you will get this wrong answer.

\[x, 16, y, 28, 34, \ldots\]

The second term of the sequence above is \(16\). For any positive integer \(n\), which of the following linear expressions could be the formula for finding the \(n^\text{th}\) term of the sequence?

(A) \(\ \ 5n+6\)
(B) \(\ \ 6n+4\)
(C) \(\ \ 6n+5\)
(D) \(\ \ 7n+3\)
(E) \(\ \ 8n\)

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Correct Answer: B

Solution 1:

Because the formula must be a linear expression, it must have the form \(an+b.\)

For \(n=5\), we have: \(a\cdot 5+b=34. \quad (1)\)
For \(n=4\), we have: \(a\cdot 4+b=28 .\quad (2)\)

Subtracting the second equation from the first, we obtain

\[\begin{array}{l c l l l} a\cdot 5+b-(a\cdot 4+b)&=&34-28 &\quad \text{subtract}\ (2)\ \text{from}\ (1)&(3)\\ 5a+b-4a-b&=&6 &\quad \text{use distributive property} &(4)\\ a&=&6. &\quad \text{combine like terms} &(5)\\ \end{array}\]

The solution must be of the form \(6n+b\). To find \(b\), notice that for \(n=2\), we have

\[\begin{array}{l c l l l} a\cdot 2+b&=&16 &\quad \text{linear expression for}\ n=2 &(6)\\ 6\cdot 2+b&=&16 &\quad \text{substitute}\ 6\ \text{for}\ a &(7)\\ 12+b&=&16 &\quad 6\cdot 2=12 &(8)\\ b&=&4. &\quad \text{subtract}\ 12\ \text{from both sides} &(9)\\ \end{array}\]

Therefore, the formula for finding the \(n^\text{th}\) term of the sequence is \(6n+4\).

Solution 2:

Tip: Plug and check.
The second term of the sequence is \(16\), the fourth term of the sequences is \(28\), and the fifth term of the sequence is \(34\). So, the formula should yield the following results:

For \(n=2\), the result should be \(16\),
For \(n=4\), the result should be \(28\),
For \(n=5\), the result should be \(34\).

We try each answer choice and find the one that doesn't lead to a contradiction.

(A)
For \(n=2\): \(\quad 5n+6=5\cdot2+6=10+6=16\).
For \(n=4\): \(\quad 5n+6=5\cdot4+6=20+6=26 \neq28\). Eliminate (A).

(B)
For \(n=2\): \(\quad 6n+4=6\cdot 2 +4 =12+4=16\).
For \(n=4\): \(\quad 6n+4=6\cdot 4+4=24+4=28\).
For \(n=5\): \(\quad 6n+4=6\cdot 5+4=30+4=34\). No contradictions.

(C)
For \(n=2\): \(\quad 6n+5=6\cdot2+5=14+2=17\neq 16\). Eliminate (C).

(D)
For \(n=2\): \(\quad 7n+3=7\cdot2+3=14+3=17\neq16\). Eliminate (D).

(E)
For \(n=2\): \(\quad 8n=8\cdot2=16\).
For \(n=4\): \(\quad 8n=8\cdot4=32 \neq 28\). Eliminate (E).

Only answer choice (B) yielded no contradiction. It is the correct answer.

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Incorrect Choices:

(A), (C), (D), (E)
Solution 1 eliminates these choices by finding the linear expression. Solution 2 eliminates them by plugging and checking.

Series are quite popular exercises solved by students, but here is a proof that a series can have any answer for the next term. I have used the series given in a question named 'Sequence' by Poo La. In the question, the series is \(1,11,21,1211,111221\) and we have to find the next term.

If you thought these examples difficult and you need to review the material, these links will help:

• Arithmetic Sequences: \(1, 3, 5, 7, \ldots, t_{n}\)
• Geometric Sequences: \(1, \frac{1}{2}, \frac{1}{4}, \ldots, t_{n}\)
• Arithmetic Series: \(1+3+5+7+\cdots +t_{n}\)
• Geometric Series: \(1 + \frac{1}{2} + \frac{1}{4} + \cdots+t_{n}\)
• SAT Exponents
• Combining Like Terms
• Distributive Property: \(a(b+c)=ab+ac\)
• Simple Equations

• Use a calculator.
• Look for shortcuts.
• SAT General Tips