Arithmetic Sequences and Series: Complete Problems with Solutions

The sequence is an arithmetic sequence because the difference between consecutive terms is constant.
It is known that $a =-1$ and $b = 2$, so
$\begin{aligned} \text{U}_n & = a + (n-1)b \\ & =-1 + (n-1) \times 2 \\ & =-1 + 2n- 2 \\ & = 2n-3. \end{aligned}$
Therefore, the general formula for the $n$-th term is $\boxed{\text{U}_n = 2n-3}.$
(Answer D)

The number sequence is an arithmetic sequence because the difference between consecutive terms is constant.
It is known that $a =-8$ and $b = 8$.
Using the formula for the $n$-th term of an arithmetic sequence, we obtain
$\begin{aligned} \text{U}_n & = a +(n-1)b \\ & =-8 + (n-1)\times 8 \\ & =-8 + 8n- 8 \\ & = 8n-16. \end{aligned}$
Therefore, the general formula of the sequence is $\boxed{\text{U}_n = 8n-16}.$
(Answer E)

The sequence is an arithmetic sequence because the difference between consecutive terms is constant. It is known that $a =-18$ and $b = 3$, so
$\begin{aligned} \text{U}_n & = a + (n-1)b \\ & =-18 + (n-1) \times 3 \\ & =-18 + 3n-3 = 3n-21. \end{aligned}$
Therefore, the general formula for the $n$-th term is $\boxed{\text{U}_n = 3n-21}.$
(Answer E)

The sequence above is an arithmetic sequence with first term $a = 5$ and $b =-3$. Thus, we obtain
$\begin{aligned} \text{U}_n & = a + (n-1)b \\ & = 5 + (n-1)(-3) \\ & = 5-3n + 3 \\ & =-3n + 8. \end{aligned}$
Therefore, the formula for the $n$-th term of the sequence is $\boxed{\text{U}_n =-3n + 8}.$
(Answer E)

The sequence is an arithmetic sequence because the difference between consecutive terms is constant.
It is known that $a = 6$ and $b = 4$, so
$\begin{aligned} \text{U}_n & = a + (n-1)b \\ & = 6 + (n-1) \times 4 \\ & = 6 + 4n- 4 \\ & = 4n + 2. \end{aligned}$
Therefore, the general formula for the $n$-th term is $\boxed{\text{U}_n = 4n + 2}.$
(Answer C)

It is known that: $a = 4$ and $b =-3$. Thus,
$\begin{aligned} \text{U}_{n} & = a + (n-1)b \\ \text{U}_{10} & = 4 + (10-1) \times (-3) \\ & = 4 + 9 \times (-3) \\ & = 4-27 =-23. \end{aligned}$
Therefore, the 10th term of the arithmetic sequence is $\boxed{-23}.$
(Answer C)

It is known that $\text{U}_n = 15-3n$. For $n = 15$, we obtain
$\text{U}_{15} = 15-3(15) = 15-45 =-30.$
Therefore, the 15th term of the sequence is $\boxed{-30}.$
(Answer E)

It is known that the formula for the $n$-th term of an arithmetic sequence is $\text{U}_n = a + (n-1)b$. We will first find the value of $b$ (common difference) as follows.
$b = \dfrac{\text{U}_9- \text{U}_5}{9-5} = \dfrac{6-18}{4} =-3$
Next, we will find the value of $a$ (first term) using the equation $\text{U}_5 = 18$ as follows.
$\begin{aligned} \text{U}_5 = a + 4b & = 18 \\ a + 4(-3) & = 18 \\ a & = 30 \end{aligned}$
The 3rd term of the sequence is $\boxed{\text{U}_3 = a + 2b = 30 + 2(-3) = 24}.$
(Answer E)

It is known that the formula for the $n$-th term of an arithmetic sequence is $\text{U}_n = a + (n-1)b$. We will first find the value of $b$ (common difference) as follows.
$b = \dfrac{\text{U}_5- \text{U}_3}{5-3} = \dfrac{-9-(-5)}{2} =-2$
Next, we will find the value of $a$ (first term) using the equation $\text{U}_3 =-5$ as follows.
$\begin{aligned} \text{U}_3 = a + 2b & =-5 \\ a + 2(-2) & =-5 \\ a-4 & =-5 \\ a & =-1 \end{aligned}$
The 10th term of the sequence is $\boxed{\text{U}_{10} = a + 9b =-1 + 9(-2) =-19}.$
(Answer D)

It is known that the formula for the $n$-th term of an arithmetic sequence is $\text{U}_n = a + (n-1)b$. We will first find the value of $b$ (common difference) as follows.
$b = \dfrac{\text{U}_9- \text{U}_4}{9-4} = \dfrac{37-17}{5} = 4$
Next, we will find the value of $a$ (first term) using the equation $\text{U}_4 = 17$ as follows.
$\begin{aligned} \text{U}_4 = a + 3b & = 17 \\ a + 3(4) & = 17 \\ a + 12 & = 17 \\ a & = 5 \end{aligned}$
The 7th term of the sequence is $\boxed{\text{U}_7 = a + 6b = 5 + 6(4) = 29}.$
(Answer B)

It is known that the formula for the $n$-th term of an arithmetic sequence is $\text{U}_n = a + (n-1)b$. We will first find the value of $b$ (common difference) as follows.
$b = \dfrac{\text{U}_5-\text{U}_1}{5-1} = \dfrac{11-3}{4} = 2$
The 25th term of the sequence is $\boxed{\text{U}_{25} = a + 24b = 3 + 24(2) = 51}.$
(Answer E)

Notice that $\text{U}_5 = 17$ and $\text{U}_{10} = 32$.
From this, we know that for every five terms, the difference is $32-17 = 15$.
Therefore,
$\text{U}_{15} = 32+15 = 47$ and $\text{U}_{20} = 47+15 = 62.$
Therefore, the 20th term of the arithmetic sequence is $\boxed{62}.$
(Answer B)

It is known that $a = 20$ and $\text{U}_6 = 40.$
The first step is to find the value of $b$ (common difference).
$\begin{aligned} \text{U}_6 & = 40 \\ a + 5b & = 40 \\ 20 + 5b & = 40 \\ 5b & = 20 \\ b & = 4 \end{aligned}$
Thus, we will calculate the value of $\text{S}_{10}$ as follows.
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{10} & = \dfrac{10}{2}\left(2 \cdot 20 + (10-1) \cdot 4\right) \\ & = 5(40 + 36) \\ & = 5(76) = 380 \end{aligned}$
Therefore, the sum of the first ten terms of the series is $\boxed{380}.$
(Answer E)

Notice that the series above is an arithmetic series because consecutive terms differ by $1$.
The number of terms in the series is
$n = 50-a + 1 = 51-a.$
It is known that $S_{n} = 1.139$, so we can write
$$\begin{aligned} S_n & = \dfrac{n} {2}(\text{U}_1 + \text{U}_n) \\ 1.139 & = \dfrac{51-a} {2}(a + 50) \\ 51a + 2.550-a^2-50a & = 2.278 \\ a^2-a-272 & = 0 \\ (a-17)(a + 16) & = 0 \end{aligned}$$Therefore, we obtain $a = 17$ or $a =-16.$
Since $a$ is a positive integer, we choose $\boxed{a=17}.$
(Answer C)

Suppose $m$ is an integer located between $\sqrt[3]{2.006}$ and $\sqrt{2.006},$ then we write $\sqrt[3]{2.006} < m < \sqrt{2.006}.$
It is known that $12^3 = 1.728$, while $13^3 = 2.197$, so the rounding up of $\sqrt[3]{2.006}$ is $13$.
It is also known that $44^2 = 1.936$ and $45^2 = 2.025$, so the rounding down of $\sqrt{2.006}$ is $44.$
Thus, we can write $13 \leq m \leq 44.$
The sum of the values of $m$ will form an arithmetic series with $a = 13, n = 44-13+1 = 32$, and $\text{U}_{32} = 44$, so
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(a + \text{U}_n) \\ \text{S}_{32} & = \dfrac{32} {2}(13 + 44) \\ & = 16 \cdot 57 = 912. \end{aligned}$
Therefore, the sum of all integers between $\sqrt[3]{2.006}$ and $\sqrt{2.006}$ is $\boxed{912}.$
(Answer B)

It is known that $\color{red} {\text{U}_3 = a + 2b = 11}.$
Since the sum of the 6th through 9th terms is $134$, we obtain
$$\begin{aligned} \text{U}_6 + \text{U}_7 +\text{U}_8 + \text{U}_9 & = 134 \\ (a + 5b) + (a + 6b) + (a + 7b) + (a + 8b) & = 134 \\ 4a + 26b & = 134 \\ (4a + 8b) + 18b & = 134 \\ 4\color{red} {(a + 2b)} + 18b & = 134 \\ 4(11) + 18b & = 134 \\ 44 + 18b & = 134 \\ 18b & = 90 \\ b & = 5. \end{aligned}$$Since $b = 5$, we get
$\begin{aligned} a + 2b = 11 & \Rightarrow a + 2(5) = 11 \\ & \Leftrightarrow a = 1. \end{aligned}$
Therefore, the first term and the common difference of the series are $\boxed{1}$ and $\boxed{5}$, respectively.
(Answer E)

It is known that $\text{U}_1 = a = 5.$
Since the tenth term is twice the fourth term, we obtain
$\begin{aligned} \text{U}_{10} & = 2\text{U}_4 \\ a + 9b & = 2(a + 3b) \\ \text{Substitute}~a&=5 \\ 5 + 9b & = 2(5+3b) \\ 5+9b&=10+6b \\ 9b-6b&=10-5 \\ 3b & = 5 \\ b & = \dfrac53. \end{aligned}$
Thus,
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_6 & = \dfrac{6}{2}\left(2 \times 5 + (6-1)\times \dfrac53\right) \\ & = 3\left(10 + \dfrac{25}{3}\right) \\ & = 30 + 25 = 55. \end{aligned}$
Therefore, the sum of the first six terms of the sequence is $\boxed{55}.$
(Answer A)

The intended arithmetic sequence is
$\begin{aligned} & 20, \text{U}_2, \text{U}_3, \text{U}_4,\text{U}_5,\text{U}_6, \\ & 68, \text{U}_8,\text{U}_9,\text{U}_{10},\text{U}_{11},\text{U}_{12}, 116 \end{aligned}$
It is known:
$\color{red} {\text{U}_1 = a = 20}.$
Since $\text{U}_7 = 68$, we obtain
$\begin{aligned}\color{red} {a} + 6b & = 68 \\ 20+6b & = 68 \\ 6b &=48 \\ b&=8 \end{aligned}$
Next, we will calculate the sum of the first 13 terms of the sequence.
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{13} & = \dfrac{13}{2}(2 \times 20 + (13-1)\times 8) \\ & = \dfrac{13}{2}(40 + 96) \\ & = \dfrac{13}{\cancel{2}} \times \cancelto{68}{136} = 884. \end{aligned}$
Therefore, the sum of all inserted numbers is
$$\boxed{\text{S}_{13}- 20- 68- 116 = 884-204 = 680}.$$(Answer A)

Recall that $\text{U}_n = \text{S}_n- \text{S}_{n-1},$
Therefore, we will calculate the values of $\text{S}_{10}$ and $\text{S}_9$ as follows.
$\begin{aligned} \text{S}_n & = \dfrac{5}{2}n^2 + \dfrac{3}{2}n \\ \text{S}_{9} & = \dfrac{5}{2}(9)^2 + \dfrac{3}{2}(9) \\ & = \dfrac{5 \times 81}{2} + \dfrac{27}{2} \\ & = \dfrac{405 + 27}{2} = 216. \end{aligned}$
$\begin{aligned} \text{S}_n & = \dfrac{5}{2}n^2 + \dfrac{3}{2}n \\ \text{S}_{10} & = \dfrac{5}{2}(10)^2 + \dfrac{3}{\cancel{2}}(\cancelto{5}{10}) \\ & = \dfrac{5}{\cancel{2}}(\cancelto{50}{100})+ 15 \\ & = 250 + 15 = 265. \end{aligned}$
Therefore, we obtain
$\text{U}_{10} = \text{S}_{10}- \text{S}_9 = 265-216 = 49.$
Therefore, the 10th term of the arithmetic series is $\boxed{49}.$
(Answer A)

Recall that $\text{U}_n = \text{S}_n- \text{S}_{n-1}.$
Thus, the values of $\text{S}_{9}$ and $\text{S}_8$ will be determined as follows.
$\begin{aligned} \text{S}_n & = 2n^2+4n \\ \text{S}_{8} & = 2(8)^2 + 4(8) \\ & = 128 + 32 = 160. \end{aligned}$
$\begin{aligned} \text{S}_n & = 2n^2+4n \\ \text{S}_9 & = 2(9)^2 + 4(9) \\ & = 162+36=198. \end{aligned}$
Therefore, we obtain
$\text{U}_{9} = \text{S}_{9}-\text{S}_8 = 198-160=38.$
So, the $9$th term of the arithmetic series is $\boxed{38}.$
(Answer C)

Known: $\text{S}_{20} = 500; \text{U}_1=a=5.$
Using the arithmetic series sum formula, we obtain
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(a + \text{U}_n) \\ \text{S}_{20} & = \dfrac{20} {2}(5 + \text{U}_{20}) \\ \cancelto{50}{500} & = \cancel{10}(5+\text{U}_{20}) \\ 50 & = 5 + \text{U}_{20} \\ \text{U}_{20} & = 50-5 = 45. \end{aligned}$
So, the last term of the series is $\boxed{45}.$
(Answer C)

Even numbers divisible by $3$ are multiples of $6$.
The sequence of multiples of $6$ from $1$ to $101$ is
$6, 12, 18, 24, \cdots, 96.$
which is an arithmetic sequence.
Known: $a=6, n = \dfrac{96}{6} = 16$, and $\text{U}_n = \text{U}_{16} = 96$.
The sum of the terms of this sequence is expressed by
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(a+\text{U}_n) \\ \text{S}_{16} & = \dfrac{16}{2}(6+96) \\ & = 8(102) = 816. \end{aligned}$
Next, the sum of even numbers from $1$ to $101$ will be determined, namely the sum of the terms of the sequence $2,4,6,8,\cdots,100$ which is an arithmetic sequence with $a=2, n = 50$, and $\text{U}_{50} = 100$ so that
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(a+\text{U}_n) \\ \text{S}_{50} & = \dfrac{50}{2}(2+100) \\ & = 25(102) = 2.550. \end{aligned}$
Therefore, the sum of even numbers from $1$ to $101$ that are not divisible by $3$ is $\boxed{\text{S} = 2.550-816 = 1.734}.$
(Answer B)

Notice that $x_{k+1} = x_k + \dfrac12$ is equivalent to $x_{k+1}-x_k = \dfrac12$. The difference between $x_{k+1}$ and $x_k$ is constant, namely $\dfrac12$ for every $k \geq 1$ so that $x_1+x_2+x_3+\cdots+x_{400}$ is an arithmetic series with first term $x_1 = a = 1$ and $b = \dfrac{1}{2}$, as well as $n = 400$.
Therefore, we obtain
$\begin{aligned} & x_1+x_2+x_3+\cdots+x_{400} \\ & = \dfrac{n} {2}(2a+(n-1)b) \\ & = \dfrac{400}{2}\left(2(1) + (400-1) \cdot \dfrac12\right) \\ & = 200\left(2 + \dfrac{399}{2}\right) \\ & = 400 + 39.900 = 40.300. \end{aligned}$
So, the result of $\boxed{x_1+x_2+x_3+\cdots+x_{400} = 40.300}.$
(Answer B)

The difference between the $n$th term and the first term is $24$, so it can be written as
$\text{U}_n- \text{U}_1 = 24 \Leftrightarrow \text{U}_n = \text{U}_1 + 24.$
Since the middle term of the arithmetic sequence is $17$, we obtain
$\begin{aligned} \dfrac{\text{U}_1 + \text{U}_n} {2} & = 17 \\ \text{U}_1 + ( \text{U}_1 + 24) & = 17 \cdot 2 \\ 2\text{U}_1 & = 10 \\ \text{U}_1 & = 5. \end{aligned}$
So, the first term of the sequence is $\boxed{5}.$
(Answer C)

Since the sum of the $3$rd and $5$th terms is $14$, we obtain
$$\begin{aligned} \text{U}_3 + \text{U}_5 & = 14 \\ (a + 2b) + (a + 4b) & = 14 \\ 2a + 6b & = 14 \\ 2a & = 14-6b && (\bigstar) \end{aligned}$$Since the sum of its first $12$ terms is $129$, we obtain
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a+(n-1)b) \\ \text{S}_{12} & = \dfrac{12}{2}((14-6b) + (12-1)b) \\ 129 & = 6(5b +14) \\ 129 & = 30b + 84 \\ 30b & = 45 \\ b & = \dfrac{45}{30} = \dfrac{3}{2}. \end{aligned}$
Substitute the value of $b = \dfrac32$ into equation $\bigstar.$
$\begin{aligned} 2a & = 14-\cancelto{3}{6}\left(\dfrac{3}{\cancel{2}} \right) \\ 2a & = 14- 9 \\ 2a & = 5 \\ a & = \dfrac52 \end{aligned}$
Since the $n$th term is $193$, we write
$\begin{aligned} \text{U}_n & = a + (n-1)b \\ 193 & = \dfrac52 + (n-1)\left(\dfrac32\right) \\ \text{Multiply}~2&~\text{on both sides} \\ 386 & = 5 + (n-1)(3) \\ 381 & = 3(n-1) \\ n-1 & = \dfrac{381}{3} = 127 \\ n & = 128. \end{aligned}$
So, the value of $n$ is $\boxed{128}.$
(Answer D)

It is known that $\text{U}_{25} = 3\text{U}_5$.
Based on the formula for the $n$th term of an arithmetic sequence, namely $\text{U}_n = a + (n-1)b$, we obtain
$\begin{aligned} a + 24b & = 3(a + 4b) \\ a + 24b & = 3a + 12b \\ 2a & = 12b \\ a & = 6b. \end{aligned}$
Suppose the term whose value is twice the first term is the $n$th term, then we write
$\begin{aligned} \text{U}_n & = 2\text{U}_1 \\ a + (n-1)b & = 2a \\ (n-1)b & = a \\ \text{Substitute}~a & = 6b \\ (n-1)\cancel{b} & = 6\cancel{b} \\ n-1 & = 6 \\ n & = 7. \end{aligned}$
Therefore, the term whose value is twice the first term is the $7$th term.
(Answer D)

Suppose the sum of the terms of the original arithmetic sequence is $\text{S}_k = \text{U}_1 + \text{U}_2 + \cdots + \text{U}_k$, while the sum of the terms of the new arithmetic sequence is $\text{S}_n$ with
$(\text{U}_1 + 3) + (\text{U}_2 + 9) + (\text{U}_3 + 15) +$ $\cdots + (\text{U}_k + x) = 1.092.$
By grouping the terms, we write
$\begin{aligned} (\text{U}_1 & + \text{U}_2 + \text{U}_3 + \cdots + \text{U}_k) + (3 \\ & + 9 + 15 + \cdots + x) = 1.092. \end{aligned}$
Thus, we obtain
$\begin{aligned} 585 + (3 + 9 + 15 + \cdots + x) & = 1.092 \\ 3 + 9 + 15 + \cdots + x & = 507. \end{aligned}$
Notice that the series $3 + 9 + 15 + \cdots + x$ is an arithmetic series with first term $a = 3$, common difference $b = 6$, and $\text{S}_n = 507.$
We will determine the value of $n$.
$\begin{aligned} \text{S}_n & = \dfrac{n} {2}(2a + (n-1)b) \\ 507 & = \dfrac{n}{2}(2 \cdot 3 + (n-1) \cdot 6) \\ 507 & = \dfrac{n} {2} (6 + 6n- 6) \\ 507 & = 3n^2 \\ n^2 & = 169 \\ n & = 13 \end{aligned}$
This means that $n = k = 13$.
The sum of the first term and the last term of the original arithmetic sequence can be determined using the formula for $\text{S}_k$.
$\begin{aligned} \text{S}_k & = \dfrac{k} {2}(\text{U}_1 + \text{U}_k) \\ 585 & = \dfrac{13} {2}(\text{U}_1 + \text{U}_{13}) \\ (\text{U}_1 + \text{U}_{13}) & = \dfrac{585 \times 2}{13} = 90 \end{aligned}$
The middle term is
$\text{U}_7 = \dfrac{\text{U}_1 + \text{U}_{13}} {2} = \dfrac{90}{2} = 45.$
Thus, the result of
$\text{U}_1 + \text{U}_7 + \text{U}_{13} = 90 + 45 = 135.$
Therefore, the sum of the first term, middle term, and last term of the sequence is $\boxed{135}.$
(Answer C)

In an arithmetic sequence, the formula for the $n$th term is given by $\text{U}_n = a + (n-1)b$.
Therefore, we obtain
$$\begin{aligned} \text{U}_{6}-\text{U}_{8}+\text{U}_{10}-\text{U}_{12}+\text{U}_{14}& =20 \\ (a+5b)-(a+7b)+(a+9b)-(a+11b)+(a+13b)&=20 \\ (a-a+a-a+a)+(5b-7b+9b-11b+13b)&=20 \\ a + 9b & = 20 \\ 2a + 18b & = 40. && (\text{Multiply by}~2) \end{aligned}$$The sum of the first $19$ terms of the arithmetic sequence is expressed by
$\begin{aligned} \text{S}_{19} & = \dfrac{19}{2}(\text{U}_1 + \text{U}_{19}) \\ & = \dfrac{19}{2}(a + (a + 18b)) \\ & = \dfrac{19}{2}(2a + 18b) \\ & = \dfrac{19}{\cancel{2}}(\cancelto{20}{40}) \\ & = 380. \end{aligned}$
Therefore, the sum of the first $19$ terms of the arithmetic sequence is $\boxed{380}.$
(Answer B)

Given $$1+2+3+\cdots+n = \overline{aaa}.$$Based on the formula for the sum of an arithmetic series (left side) and the expansion of a three-digit number (right side), we obtain
$$\begin{aligned} \dfrac{n}{2}(1+n) & = 100a+10a+a \\ n(n+1) & = 222a \\ n(n+1) & = 2 \cdot 3 \cdot 37a \\ n(n+1) & = 6 \cdot 37a. \end{aligned}$$The value of $a$ that produces the multiplication of two consecutive integers (as on the left side) is $a = 6$ so that
$$\begin{aligned} n(n+1) & = 6 \cdot 37(6) \\ n(n+1) & = 36 \cdot 37. \end{aligned}$$Thus, the values obtained are $n = 36$ and $\overline{aaa} = 666.$