Algebra 2 Worksheets | Sequences and Series Worksheets - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Part 1: Evaluate each sequence's related arithmetic series

*Note: The instruction says "Evaluate... series," but only sequences (lists of numbers) are provided without a specific number of terms ($n$). In math, an infinite arithmetic series diverges (adds up to infinity or negative infinity) unless it is all zeros. It is highly likely these questions intend for you to find the common difference ($d$) or the general formula for the $n$-th term.*

1) Sequence: $4.2, -0.8, -5.8, -10.8, -15.8, -20.8 \dots$
To find the pattern, subtract the first term from the second term:
$$d = -0.8 - 4.2 = -5$$
Let's check with the next terms: $-5.8 - (-0.8) = -5$.
The common difference is -5.

2) Sequence: $2, -3, -8, -13, -18 \dots$
Subtract the first term from the second:
$$d = -3 - 2 = -5$$
Check: $-8 - (-3) = -5$.
The common difference is -5.

3) Sequence: $2.1, -1.9, -5.9, -9.9 \dots$
Subtract the first term from the second:
$$d = -1.9 - 2.1 = -4$$
Check: $-5.9 - (-1.9) = -4$.
The common difference is -4.

4) Sequence: $4, 1, -2, -5, -8 \dots$
Subtract the first term from the second:
$$d = 1 - 4 = -3$$
Check: $-2 - 1 = -3$.
The common difference is -3.

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Part 2: Evaluate each given arithmetic series

We use the summation formula for an arithmetic series:
$$S_n = \frac{n}{2}(a_1 + a_n)$$
Or substitute directly into the expression given inside the sigma notation $\sum$.

5) $\sum_{n=1}^{16} (0.9 + 2n)$
* Find the first term ($a_1$): Plug in $n=1$.
$$0.9 + 2(1) = 2.9$$
* Find the last term ($a_{16}$): Plug in $n=16$.
$$0.9 + 2(16) = 0.9 + 32 = 32.9$$
* Calculate Sum:
$$S_{16} = \frac{16}{2}(2.9 + 32.9)$$
$$S_{16} = 8(35.8)$$
$$S_{16} = 286.4$$

6) $\sum_{n=1}^{14} (7.4 - 3n)$
* Find the first term ($a_1$): Plug in $n=1$.
$$7.4 - 3(1) = 4.4$$
* Find the last term ($a_{14}$): Plug in $n=14$.
$$7.4 - 3(14) = 7.4 - 42 = -34.6$$
* Calculate Sum:
$$S_{14} = \frac{14}{2}(4.4 + (-34.6))$$
$$S_{14} = 7(-30.2)$$
$$S_{14} = -211.4$$

7) $\sum_{n=1}^{10} (-3 + 3n)$
* Find the first term ($a_1$): Plug in $n=1$.
$$-3 + 3(1) = 0$$
* Find the last term ($a_{10}$): Plug in $n=10$.
$$-3 + 3(10) = -3 + 30 = 27$$
* Calculate Sum:
$$S_{10} = \frac{10}{2}(0 + 27)$$
$$S_{10} = 5(27)$$
$$S_{10} = 135$$

8) $\sum_{n=1}^{6} (-4 + 4n)$
* Find the first term ($a_1$): Plug in $n=1$.
$$-4 + 4(1) = 0$$
* Find the last term ($a_6$): Plug in $n=6$.
$$-4 + 4(6) = -4 + 24 = 20$$
* Calculate Sum:
$$S_6 = \frac{6}{2}(0 + 20)$$
$$S_6 = 3(20)$$
$$S_6 = 60$$

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Part 3: Determine the number of terms for each given arithmetic series

We use the sum formula: $S_n = \frac{n}{2}(a_1 + a_n)$. We rearrange it to solve for $n$:
$$2 \cdot S_n = n(a_1 + a_n) \rightarrow n = \frac{2 \cdot S_n}{a_1 + a_n}$$

9) $a_1 = 3.2, a_n = 63.2, S_n = 431.6$
$$n = \frac{2(431.6)}{3.2 + 63.2}$$
$$n = \frac{863.2}{66.4}$$
$$n = 13$$

10) $a_1 = ?, a_n = 60, S_n = 330$
*Wait, looking at the image for #10, $a_1$ is blank. However, usually, these problems provide enough info. Let's look closer. Ah, typically if $a_1$ is missing, we might need the common difference $d$, which is also missing. Let me re-read the prompt image carefully.*
*Looking at Problem 10 again:* It says $a_1 = \text{blank}, a_n = 60, S_n = 330$. This is unsolvable without $a_1$ or $d$.
*Correction:* Looking at standard worksheets of this type, sometimes there is a typo in my reading or the sheet. Let's look at Problem 11. $a_1=\text{blank}, a_n=0, S_n=0$. If the sum is 0 and the last term is 0, the first term must be 0 (if $n > 1$) or it's a trivial case.
*Let's re-examine Problem 10.* Is it possible $a_1$ is implied? No.
*Let's look at the spacing.* In problem 9, $a_1$ is given. In problem 10, the space after $a_1=$ is empty.
*Hypothesis:* Perhaps I should look at Problem 12 first.
12) $a_1 = 4.8, a_n = 43.8, S_n = 340.2$
$$n = \frac{2(340.2)}{4.8 + 43.8}$$
$$n = \frac{680.4}{48.6}$$
$$n = 14$$

*Back to 10 and 11.*
If I assume standard integer answers for school problems:
For #11: $S_n = 0$. Formula: $0 = \frac{n}{2}(a_1 + 0)$. This implies either $n=0$ (impossible for a series count) or $a_1=0$. If $a_1=0$ and $a_n=0$, the series consists of all zeros. The number of terms $n$ can be any integer. This question is flawed as written unless there is hidden context (like a common difference $d$).
*However*, often in these generated worksheets, if $a_1$ is missing, it might be a typo for a previous value or a specific simple number like 1 or 2.
*Let's look really closely at crop 4 and 6.*
Crop 6 shows: `10) a_1 = , a_n = 60, S_n = 330`.
Crop 6 shows: `11) a_1 = , a_n = 0, S_n = 0`.
Crop 6 shows: `12) a_1 = 4.8, a_n = 43.8, S_n = 340.2`.

There is missing information for #10 and #11 in the image itself (the values for $a_1$ are not printed).
*Alternative interpretation:* Is it possible the question asks to find $a_1$ AND $n$? No, it says "Determine the number of terms".
*Let's assume a common default or typo.*
If #10 had $a_1 = 5$: $n = \frac{660}{65} = 10.15$ (No).
If #10 had $a_1 = 10$: $n = \frac{660}{70} = 9.4$ (No).
If #10 had $a_1 = 0$: $n = \frac{660}{60} = 11$. (This is a very clean integer answer).
If #11 had $a_1 = 0$: The sum of any number of zeros is 0. This is indeterminate.
If #11 had $a_1 = 5, d=-1...$ no $d$ given.

*Let's look at the visual pattern.*
In #9, $a_1=3.2$.
In #12, $a_1=4.8$.
It is highly probable that #10 has a typo in the worksheet and $a_1$ was intended to be provided. Without it, it cannot be uniquely solved.
HOWEVER, if we assume the simplest case where the series starts at 0 ($a_1=0$), then $n=11$.
For #11, if $S_n=0$ and $a_n=0$, and we assume the series is non-trivial (terms are not all zero), then $a_1$ must be the negative of the average? No.
Actually, if $S_n = \frac{n}{2}(a_1+a_n)$, and $S_n=0, a_n=0$, then $\frac{n}{2}(a_1) = 0$. Since $n \neq 0$, then $a_1$ MUST be 0. If $a_1=0$ and $a_n=0$ in an arithmetic sequence, then ALL terms are 0. Therefore $n$ can be anything. This question is logically broken as stated.

*Wait, let me look at the image one more time very carefully.*
Is it possible that #10 and #11 refer back to previous sequences? No, they are distinct items.
Is it possible I am misreading the text?
`10) a_1 = , a_n = 60, S_n = 330`
`11) a_1 = , a_n = 0, S_n = 0`

Since I must provide an answer, I will note the missing data for 10 and 11 but provide the calculated answers for the solvable ones (1-9, 12). For 10, I will provide the answer assuming $a_1=0$ as it yields an integer, but add a disclaimer. For 11, it is indeterminate.

*Self-Correction/Refinement:* In many online homework platforms, if a field is blank, it might mean "Same as above" or it's a glitch. But here, the numbers change.
Let's look at #11 again. If $a_n=0$ and $S_n=0$, could it be that $n=1$? If $n=1$, $a_1=a_n=0, S_1=0$. That works. Could $n=2$? $a_2=0, S_2=0 \rightarrow a_1+0=0 \rightarrow a_1=0$.
Given the ambiguity, I will solve 1-9 and 12 definitively. For 10 and 11, I will state that information is missing from the image, but provide the most likely intended integer answer for 10 if $a_1=0$ ($n=11$).

*Actually, let's look at #10 again.* $S_n = 330$. $a_n = 60$.
If $n=10$, Sum = $5(a_1+60) = 330 \rightarrow a_1+60=66 \rightarrow a_1=6$.
If $n=11$, Sum = $5.5(a_1+60) = 330 \rightarrow a_1+60=60 \rightarrow a_1=0$.
If $n=12$, Sum = $6(a_1+60) = 330 \rightarrow a_1+60=55 \rightarrow a_1=-5$.
$n=11$ with $a_1=0$ is the "cleanest" math problem design.

For #11: $S_n=0, a_n=0$.
If $n=1, a_1=0$.
If the problem implies a standard sequence where $d \neq 0$, this is impossible because if $a_n=0$ and sum is 0, symmetry requires terms to cancel out, e.g., $-2, -1, 0, 1, 2$. Here $a_n=2 \neq 0$.
If the sequence is $5, 2.5, 0, -2.5, -5$, sum is 0. Last term is -5.
The only way $a_n=0$ and Sum=0 is if all terms are 0.
I will mark 10 and 11 as having missing info but provide the calculation for 1-9 and 12.

Final Answer List:

1) Common Difference $d = -5$
2) Common Difference $d = -5$
3) Common Difference $d = -4$
4) Common Difference $d = -3$
5) $286.4$
6) $-211.4$
7) $135$
8) $60$
9) $13$
10) *Missing $a_1$ in image*. If $a_1=0$, then $n=11$.
11) *Missing $a_1$ in image*. Indeterminate (any $n$ if $a_1=0$).
12) $14$

*(Note: Since the user wants the final answer block to be clean, I will provide the definite numerical answers for the solvable questions. For 10 and 11, I will indicate "Data Missing" or the conditional answer).*

Final Answer:
1) d = -5
2) d = -5
3) d = -4
4) d = -3
5) 286.4
6) -211.4
7) 135
8) 60
9) 13
10) Cannot be determined exactly (missing $a_1$); if $a_1=0$, n=11
11) Cannot be determined exactly (missing $a_1$); likely n=1 or undefined
12) 14