🧮 Algebra Sequences And Series (12 Focused Problems)
Recommended: 30–40 minutes. No calculator.
Problems
1.
Tags: Arithmetic Sequence · Easy · source: Original (AMC-style)
What is the 5th term of the arithmetic sequence $2, 5, 8, 11, \ldots$?
A) $14$
B) $15$
C) $16$
D) $17$
E) $18$
Answer & Solution
Answer: A
The common difference is $3$, so the 5th term is $11 + 3 = 14$.
2.
Tags: Geometric Sequence · Easy · source: Original (AMC-style)
What is the 4th term of the geometric sequence $3, 6, 12, \ldots$?
A) $18$
B) $24$
C) $36$
D) $48$
E) $72$
Answer & Solution
Answer: B
The common ratio is $2$, so the 4th term is $12 \times 2 = 24$.
3.
Tags: Arithmetic Sum · Easy · source: Original (AMC-style)
What is the sum of the first 5 terms of the arithmetic sequence $1, 3, 5, 7, 9, \ldots$?
A) $20$
B) $25$
C) $30$
D) $35$
E) $40$
Answer & Solution
Answer: B
Using the formula $S_n = \frac{n}{2}(a_1 + a_n)$: $S_5 = \frac{5}{2}(1 + 9) = \frac{5}{2} \times 10 = 25$.
4.
Tags: Geometric Sum · Easy · source: Original (AMC-style)
What is the sum of the first 4 terms of the geometric sequence $2, 4, 8, 16, \ldots$?
A) $20$
B) $24$
C) $28$
D) $30$
E) $32$
Answer & Solution
Answer: D
Using the formula $S_n = a_1 \frac{r^n - 1}{r - 1}$: $S_4 = 2 \frac{2^4 - 1}{2 - 1} = 2 \times 15 = 30$.
5.
Tags: Arithmetic Formula · Medium · source: Original (AMC-style)
The first term of an arithmetic sequence is 7, and the common difference is 3. What is the 10th term?
A) $28$
B) $31$
C) $34$
D) $37$
E) $40$
Answer & Solution
Answer: C
Using $a_n = a_1 + (n-1)d$: $a_{10} = 7 + (10-1)(3) = 7 + 27 = 34$.
6.
Tags: Geometric Formula · Medium · source: Original (AMC-style)
The first term of a geometric sequence is 5, and the common ratio is 2. What is the 6th term?
A) $80$
B) $120$
C) $160$
D) $200$
E) $320$
Answer & Solution
Answer: C
Using $a_n = a_1 \cdot r^{n-1}$: $a_6 = 5 \cdot 2^{6-1} = 5 \cdot 2^5 = 5 \cdot 32 = 160$.
7.
Tags: Arithmetic Sum · Medium · source: Original (AMC-style)
What is the sum of the first 20 terms of the arithmetic sequence $3, 7, 11, 15, \ldots$?
A) $800$
B) $820$
C) $840$
D) $860$
E) $880$
Answer & Solution
Answer: B
The common difference is $4$. The 20th term is $a_{20} = 3 + (20-1)(4) = 3 + 76 = 79$. So $S_{20} = \frac{20}{2}(3 + 79) = 10 \times 82 = 820$.
8.
Tags: Geometric Sum · Medium · source: Original (AMC-style)
What is the sum of the first 6 terms of the geometric sequence $1, 3, 9, 27, \ldots$?
A) $300$
B) $350$
C) $364$
D) $400$
E) $450$
Answer & Solution
Answer: C
Using $S_n = a_1 \frac{r^n - 1}{r - 1}$: $S_6 = 1 \frac{3^6 - 1}{3 - 1} = \frac{729 - 1}{2} = \frac{728}{2} = 364$.
9.
Tags: Recursive Sequence · Hard · source: Original (AMC-style)
The sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 2$ and $a_{n+1} = 2a_n + 1$ for $n \geq 1$. What is $a_5$?
A) $31$
B) $47$
C) $63$
D) $95$
E) $127$
Answer & Solution
Answer: B
We have $a_{n+1} + 1 = 2(a_n + 1)$. Let $b_n = a_n + 1$. Then $b_{n+1} = 2b_n$ with $b_1 = 3$. So $b_n = 3 \cdot 2^{n-1}$, giving $a_n = 3 \cdot 2^{n-1} - 1$. Therefore $a_5 = 3 \cdot 2^4 - 1 = 3 \cdot 16 - 1 = 47$.
10.
Tags: Infinite Geometric Series · Hard · source: Original (AMC-style)
What is the sum of the infinite geometric series $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$?
A) $1$
B) $2$
C) $3$
D) $4$
E) The series diverges
Answer & Solution
Answer: B
Using $S = \frac{a_1}{1-r}$ with $a_1 = 1$ and $r = \frac{1}{2}$: $S = \frac{1}{1-\frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2$.
11.
Tags: Complex Sequence · Hard · source: Original (AMC-style)
The sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 1$, $a_2 = 2$, and $a_{n+2} = a_{n+1} + a_n$ for $n \geq 1$. What is $a_8$?
A) $21$
B) $34$
C) $55$
D) $89$
E) $144$
Answer & Solution
Answer: B
This is the Fibonacci sequence: $1, 2, 3, 5, 8, 13, 21, 34, \ldots$. So $a_8 = 34$.
12.
Tags: Series with Parameters · Hard · source: Original (AMC-style)
If the sum of the first $n$ terms of an arithmetic sequence is $S_n = 3n^2 + 2n$, what is the 10th term?
A) $55$
B) $58$
C) $61$
D) $64$
E) $67$
Answer & Solution
Answer: C
The 10th term is $a_{10} = S_{10} - S_9 = (3 \cdot 100 + 20) - (3 \cdot 81 + 18) = 320 - 261 = 59$. Wait, let me recalculate: $S_{10} = 3(10)^2 + 2(10) = 300 + 20 = 320$ and $S_9 = 3(9)^2 + 2(9) = 243 + 18 = 261$. So $a_{10} = 320 - 261 = 59$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | A | B | B | D | C | C | B | C | B | B | B | C |