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Arithmetic Sequence worksheet

Grades 9–11 · CCSS HSF.BF.A.2, HSF.IF.A.3

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Find: The nth term and the sum
- First term (a₁): 2
- Common difference (d): 4
- Term number (n): 9
Find: Term 9 (aₙ)
Answer
2.
Problem 2.
- Find: The nth term and the sum
- First term (a₁): 3
- Common difference (d): 4
- Term number (n): 12
Find: Term 12 (aₙ)
Answer
3.
Problem 3.
- Find: The nth term and the sum
- First term (a₁): 2
- Common difference (d): 3
- Term number (n): 11
Find: Term 11 (aₙ)
Answer
4.
Problem 4.
- Find: The nth term and the sum
- First term (a₁): 4
- Common difference (d): 4
- Term number (n): 10
Find: Term 10 (aₙ)
Answer
5.
Problem 5.
- Find: The nth term and the sum
- First term (a₁): 3
- Common difference (d): 4
- Term number (n): 10
Find: Term 10 (aₙ)
Answer
6.
Problem 6.
- Find: The nth term and the sum
- First term (a₁): 3
- Common difference (d): 3
- Term number (n): 9
Find: Term 9 (aₙ)
Answer
7.
Problem 7.
- Find: The nth term and the sum
- First term (a₁): 4
- Common difference (d): 3
- Term number (n): 12
Find: Term 12 (aₙ)
Answer
8.
Problem 8.
- Find: The nth term and the sum
- First term (a₁): 4
- Common difference (d): 3
- Term number (n): 11
Find: Term 11 (aₙ)
Answer
9.
Problem 9.
- Find: The nth term and the sum
- First term (a₁): 3
- Common difference (d): 4
- Term number (n): 11
Find: Term 11 (aₙ)
Answer
10.
Problem 10.
- Find: The nth term and the sum
- First term (a₁): 4
- Common difference (d): 4
- Term number (n): 11
Find: Term 11 (aₙ)
Answer

Answer key: Arithmetic Sequence worksheet

1. Term 9 (aₙ): 34
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₉$ $= 2 + (9 - 1) \times 4$ $= 34$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₉$ $= \frac{9}{2} \times (2 + 34)$ $= 162$
2. Term 12 (aₙ): 47
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₂$ $= 3 + (12 - 1) \times 4$ $= 47$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₂$ $= \frac{12}{2} \times (3 + 47)$ $= 300$
3. Term 11 (aₙ): 32
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₁$ $= 2 + (11 - 1) \times 3$ $= 32$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₁$ $= \frac{11}{2} \times (2 + 32)$ $= 187$
4. Term 10 (aₙ): 40
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₀$ $= 4 + (10 - 1) \times 4$ $= 40$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₀$ $= \frac{10}{2} \times (4 + 40)$ $= 220$
5. Term 10 (aₙ): 39
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₀$ $= 3 + (10 - 1) \times 4$ $= 39$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₀$ $= \frac{10}{2} \times (3 + 39)$ $= 210$
6. Term 9 (aₙ): 27
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₉$ $= 3 + (9 - 1) \times 3$ $= 27$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₉$ $= \frac{9}{2} \times (3 + 27)$ $= 135$
7. Term 12 (aₙ): 37
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₂$ $= 4 + (12 - 1) \times 3$ $= 37$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₂$ $= \frac{12}{2} \times (4 + 37)$ $= 246$
8. Term 11 (aₙ): 34
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₁$ $= 4 + (11 - 1) \times 3$ $= 34$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₁$ $= \frac{11}{2} \times (4 + 34)$ $= 209$
9. Term 11 (aₙ): 43
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₁$ $= 3 + (11 - 1) \times 4$ $= 43$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₁$ $= \frac{11}{2} \times (3 + 43)$ $= 253$
10. Term 11 (aₙ): 44
1. To reach position n, start at the first term (a₁) and add the step d once for each of the n − 1 moves after it
  $aₙ$ $= a₁ + (n - 1) d$
2. Put in this sequence's first term, the position you want and the step size
  $a₁₁$ $= 4 + (11 - 1) \times 4$ $= 44$
3. For the total (Sₙ), multiply the number of terms by the average of the first and last terms
  $Sₙ$ $= \frac{n}{2} \times (a₁ + aₙ)$
4. Put in the number of terms, the first term and the last term
  $S₁₁$ $= \frac{11}{2} \times (4 + 44)$ $= 264$