35. Finding nth terms problems

35. Finding nth terms problems - step by step solution procedure and detailed explanation for each problem.

Below are 10 problems focused on finding the nth term of various sequences, including arithmetic sequences, geometric sequences, and other types of sequences. Each problem includes a step-by-step solution procedure and a detailed explanation to clarify the reasoning, formulas, and techniques used. These problems are designed to cover a range of scenarios, from straightforward to more complex, to help you understand how to determine the nth term in different contexts.

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Problem 1: Arithmetic Sequence (Basic)

Problem: Find the 10th term of the arithmetic sequence: 3, 7, 11, 15, ...

Solution Procedure:

1. Identify the sequence type: The sequence is arithmetic because the difference between consecutive terms is constant.
2. Find the common difference (( d )):
   • 7 - 3 = 4

     ,

     11 - 7 = 4

     ,

     15 - 11 = 4

     . So,

     d = 4

     .
3. Identify the first term (

   a_1

   ): The first term is 3.
4. Use the nth term formula for an arithmetic sequence:

   a_n = a_1 + (n-1)d

   .
5. Substitute values: For

   n = 10

   ,

   a_1 = 3

   ,

   d = 4

   :
   • a_{10} = 3 + (10-1) \cdot 4 = 3 + 9 \cdot 4 = 3 + 36 = 39

     .
6. State the answer: The 10th term is 39.

Detailed Explanation:

• Arithmetic sequence: Each term is obtained by adding a constant difference to the previous term.
• Formula derivation: The nth term is the first term plus

  (n-1)

  times the common difference, accounting for the number of steps from the first term.
• Verification: List terms: 3, 7, 11, 15, 19, 23, 27, 31, 35, 39. The 10th term is indeed 39.
• Formula:

  a_n = a_1 + (n-1)d

  .

Answer: 39.

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Problem 2: Geometric Sequence (Basic)

Problem: Find the 8th term of the geometric sequence: 2, 6, 18, 54, ...

Solution Procedure:

1. Identify the sequence type: The sequence is geometric because each term is multiplied by a constant ratio.
2. Find the common ratio (( r )):
   • \frac{6}{2} = 3

     ,

     \frac{18}{6} = 3

     ,

     \frac{54}{18} = 3

     . So,

     r = 3

     .
3. Identify the first term (

   a_1

   ): The first term is 2.
4. Use the nth term formula for a geometric sequence:

   a_n = a_1 \cdot r^{n-1}

   .
5. Substitute values: For

   n = 8

   ,

   a_1 = 2

   ,

   r = 3

   :
   • a_8 = 2 \cdot 3^{8-1} = 2 \cdot 3^7

     .
   • Calculate

     3^7 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 2187

     .
   • a_8 = 2 \cdot 2187 = 4374

     .
6. State the answer: The 8th term is 4374.

Detailed Explanation:

• Geometric sequence: Each term is the previous term multiplied by a constant ratio.
• Formula derivation: The nth term is the first term multiplied by the ratio raised to the power of

  n-1

  , reflecting the number of multiplications.
• Verification: Continue the sequence: 2, 6, 18, 54, 162, 486, 1458, 4374. The 8th term matches.
• Formula:

  a_n = a_1 \cdot r^{n-1}

  .

Answer: 4374.

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Problem 3: Arithmetic Sequence with Given Terms

Problem: The 3rd term of an arithmetic sequence is 11, and the 7th term is 23. Find the 15th term.

Solution Procedure:

1. Use the arithmetic nth term formula:

   a_n = a_1 + (n-1)d

   .
2. Set up equations using given terms:
   • For

     n = 3

     :

     a_3 = a_1 + (3-1)d = a_1 + 2d = 11

     .
   • For

     n = 7

     :

     a_7 = a_1 + (7-1)d = a_1 + 6d = 23

     .
3. Solve the system of equations:
   • Subtract the first equation from the second:

     (a_1 + 6d) - (a_1 + 2d) = 23 - 11

     .
   • Simplify:

     4d = 12

     , so

     d = 3

     .
   • Substitute

     d = 3

     into the first equation:

     a_1 + 2 \cdot 3 = 11

     , so

     a_1 + 6 = 11

     ,

     a_1 = 5

     .
4. Find the 15th term:

   a_{15} = a_1 + (15-1)d = 5 + 14 \cdot 3 = 5 + 42 = 47

   .
5. State the answer: The 15th term is 47.

Detailed Explanation:

• Why two equations? Two unknowns (

  a_1

  and ( d )) require two equations to solve.
• System solving: Subtracting eliminates

  a_1

  , allowing us to find ( d ), then back-substitute.
• Verification: Check terms:

  a_3 = 5 + 2 \cdot 3 = 11

  ,

  a_7 = 5 + 6 \cdot 3 = 23

  . Sequence: 5, 8, 11, ..., with 15th term 47.
• Formula:

  a_n = a_1 + (n-1)d

  .

Answer: 47.

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Problem 4: Geometric Sequence with Given Terms

Problem: The 2nd term of a geometric sequence is 8, and the 5th term is 64. Find the 10th term.

Solution Procedure:

1. Use the geometric nth term formula:

   a_n = a_1 \cdot r^{n-1}

   .
2. Set up equations using given terms:
   • For

     n = 2

     :

     a_2 = a_1 \cdot r^{2-1} = a_1 r = 8

     .
   • For

     n = 5

     :

     a_5 = a_1 \cdot r^{5-1} = a_1 r^4 = 64

     .
3. Solve for the common ratio (( r )):
   • Divide the second equation by the first:

     \frac{a_1 r^4}{a_1 r} = \frac{64}{8}

     .
   • Simplify:

     r^3 = 8

     , so

     r = \sqrt[3]{8} = 2

     .
4. Find the first term (

   a_1

   ): Using

   a_1 r = 8

   , substitute

   r = 2

   :

   a_1 \cdot 2 = 8

   , so

   a_1 = 4

   .
5. Find the 10th term:

   a_{10} = a_1 \cdot r^{10-1} = 4 \cdot 2^9

   .
   • Calculate

     2^9 = 512

     .
   • a_{10} = 4 \cdot 512 = 2048

     .
6. State the answer: The 10th term is 2048.

Detailed Explanation:

• Geometric properties: The ratio between terms is constant, and dividing terms eliminates

  a_1

  .
• Solving for ( r ): The exponent difference (4–1 = 3) gives

  r^3

  , and we assume a positive ratio (though

  r = -2

  could be checked if needed).
• Verification: Check:

  a_2 = 4 \cdot 2 = 8

  ,

  a_5 = 4 \cdot 2^4 = 4 \cdot 16 = 64

  . Sequence: 4, 8, 16, 32, 64, ..., with 10th term 2048.
• Formula:

  a_n = a_1 \cdot r^{n-1}

  .

Answer: 2048.

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Problem 5: Quadratic Sequence

Problem: Find the nth term of the sequence: 1, 4, 10, 19, 31, ...

Solution Procedure:

1. Check for arithmetic or geometric: Differences: 3, 6, 9, 12 (not constant, so not arithmetic). Ratios are not constant (not geometric).
2. Calculate second differences:
   • First differences: 3, 6, 9, 12.
   • Second differences:

     6-3 = 3

     ,

     9-6 = 3

     ,

     12-9 = 3

     . Constant second differences suggest a quadratic sequence.
3. Assume a quadratic form:

   a_n = an^2 + bn + c

   .
4. Set up equations using the first three terms:
   • For

     n = 1

     :

     a(1)^2 + b(1) + c = a + b + c = 1

     .
   • For

     n = 2

     :

     a(2)^2 + b(2) + c = 4a + 2b + c = 4

     .
   • For

     n = 3

     :

     a(3)^2 + b(3) + c = 9a + 3b + c = 10

     .
5. Solve the system:
   • Subtract first from second:

     (4a + 2b + c) - (a + b + c) = 4 - 1

     , so

     3a + b = 3

     .
   • Subtract second from third:

     (9a + 3b + c) - (4a + 2b + c) = 10 - 4

     , so

     5a + b = 6

     .
   • Subtract the two new equations:

     (5a + b) - (3a + b) = 6 - 3

     , so

     2a = 3

     ,

     a = \frac{3}{2}

     .
   • Substitute

     a = \frac{3}{2}

     into

     3a + b = 3

     :

     3 \cdot \frac{3}{2} + b = 3

     ,

     \frac{9}{2} + b = 3

     ,

     b = 3 - \frac{9}{2} = -\frac{3}{2}

     .
   • Substitute into

     a + b + c = 1

     :

     \frac{3}{2} - \frac{3}{2} + c = 1

     , so

     c = 1

     .
6. Write the nth term:

   a_n = \frac{3}{2}n^2 - \frac{3}{2}n + 1

   .
7. Simplify (optional):

   a_n = \frac{3n^2 - 3n + 2}{2}

   .
8. Verify: Check

   n = 4

   :

   \frac{3 \cdot 16 - 3 \cdot 4 + 2}{2} = \frac{48 - 12 + 2}{2} = 19

   , matches. For

   n = 5

   :

   \frac{3 \cdot 25 - 3 \cdot 5 + 2}{2} = \frac{75 - 15 + 2}{2} = 31

   , matches.
9. State the answer: The nth term is

   \frac{3}{2}n^2 - \frac{3}{2}n + 1

   .

Detailed Explanation:

• Quadratic sequence: Constant second differences indicate a quadratic formula.
• System of equations: Three terms provide three equations to solve for ( a, b, c ).
• Verification: Checking additional terms ensures the formula is correct.
• Formula: General quadratic form

  an^2 + bn + c

  .

Answer:

a_n = \frac{3}{2}n^2 - \frac{3}{2}n + 1

.

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Problem 6: Sequence with a Pattern (Triangular Numbers)

Problem: Find the nth term of the sequence: 1, 3, 6, 10, 15, ... (triangular numbers).

Solution Procedure:

1. Recognize the pattern: These are triangular numbers, where the nth term is the sum of the first ( n ) natural numbers.
2. Use the known formula: The nth triangular number is

   a_n = \frac{n(n+1)}{2}

   .
3. Verify:
   • n = 1

     :

     \frac{1 \cdot 2}{2} = 1

     .
   • n = 2

     :

     \frac{2 \cdot 3}{2} = 3

     .
   • n = 3

     :

     \frac{3 \cdot 4}{2} = 6

     .
   • n = 4

     :

     \frac{4 \cdot 5}{2} = 10

     .
   • n = 5

     :

     \frac{5 \cdot 6}{2} = 15

     .
4. State the answer: The nth term is

   \frac{n(n+1)}{2}

   .

Detailed Explanation:

• Triangular numbers: Represent the number of dots in a triangular array, equivalent to

  1 + 2 + \cdots + n

  .
• Formula derivation: The sum of the first ( n ) natural numbers is

  \frac{n(n+1)}{2}

  , derived from pairing terms or Gaussian summation.
• Quadratic nature: The formula is quadratic, consistent with second differences being constant (differences: 2, 3, 4, 5; second differences: 1).
• Formula:

  a_n = \frac{n(n+1)}{2}

  .

Answer:

\frac{n(n+1)}{2}

.

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Problem 7: Arithmetic Sequence with Negative Difference

Problem: Find the 12th term of the arithmetic sequence: 20, 17, 14, 11, ...

Solution Procedure:

1. Identify the sequence type: Arithmetic, as the difference is constant.
2. Find the common difference:

   17 - 20 = -3

   ,

   14 - 17 = -3

   , so

   d = -3

   .
3. Identify the first term:

   a_1 = 20

   .
4. Use the nth term formula:

   a_n = a_1 + (n-1)d

   .
5. Substitute for

   n = 12

   :

   a_{12} = 20 + (12-1) \cdot (-3) = 20 + 11 \cdot (-3) = 20 - 33 = -13

   .
6. State the answer: The 12th term is -13.

Detailed Explanation:

• Negative difference: The sequence decreases, but the arithmetic formula applies the same way.
• Verification: List terms: 20, 17, 14, 11, 8, 5, 2, -1, -4, -7, -10, -13. The 12th term is -13.
• Sign handling: The negative difference reduces the term value as ( n ) increases.
• Formula:

  a_n = a_1 + (n-1)d

  .

Answer: -13.

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Problem 8: Geometric Sequence with Fractional Ratio

Problem: Find the 7th term of the geometric sequence: 16, 8, 4, 2, ...

Solution Procedure:

1. Identify the sequence type: Geometric, as each term is multiplied by a constant ratio.
2. Find the common ratio:

   \frac{8}{16} = \frac{1}{2}

   ,

   \frac{4}{8} = \frac{1}{2}

   , so

   r = \frac{1}{2}

   .
3. Identify the first term:

   a_1 = 16

   .
4. Use the nth term formula:

   a_n = a_1 \cdot r^{n-1}

   .
5. Substitute for

   n = 7

   :

   a_7 = 16 \cdot \left(\frac{1}{2}\right)^{7-1} = 16 \cdot \left(\frac{1}{2}\right)^6

   .
6. Calculate:

   \left(\frac{1}{2}\right)^6 = \frac{1}{2^6} = \frac{1}{64}

   , so

   a_7 = 16 \cdot \frac{1}{64} = \frac{16}{64} = \frac{1}{4}

   .
7. State the answer: The 7th term is

   \frac{1}{4}

   .

Detailed Explanation:

• Fractional ratio: A ratio less than 1 causes the terms to decrease.
• Exponent rule:

  \left(\frac{1}{2}\right)^6 = \frac{1}{2^6}

  , simplifying the calculation.
• Verification: Sequence: 16, 8, 4, 2, 1,

  \frac{1}{2}

  ,

  \frac{1}{4}

  . The 7th term is

  \frac{1}{4}

  .
• Formula:

  a_n = a_1 \cdot r^{n-1}

  .

Answer:

\frac{1}{4}

.

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Problem 9: Fibonacci-Like Sequence

Problem: A sequence is defined by

a_1 = 1

,

a_2 = 3

, and

a_n = a_{n-1} + a_{n-2}

for

n \geq 3

. Find the 6th term.

Solution Procedure:

1. Identify the sequence type: This is a recursive sequence similar to the Fibonacci sequence.
2. Use the recurrence relation to compute terms:
   • a_1 = 1

     .
   • a_2 = 3

     .
   • a_3 = a_2 + a_1 = 3 + 1 = 4

     .
   • a_4 = a_3 + a_2 = 4 + 3 = 7

     .
   • a_5 = a_4 + a_3 = 7 + 4 = 11

     .
   • a_6 = a_5 + a_4 = 11 + 7 = 18

     .
3. State the answer: The 6th term is 18.

Detailed Explanation:

• Recursive sequence: Each term depends on the previous two, requiring iterative calculation.
• No closed formula: Unlike arithmetic or geometric sequences, Fibonacci-like sequences typically require computing each term unless a closed form (e.g., Binet’s formula) is derived, which is complex here.
• Verification: Sequence: 1, 3, 4, 7, 11, 18. The recurrence holds.
• Formula:

  a_n = a_{n-1} + a_{n-2}

  .

Answer: 18.

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Problem 10: Sequence with Alternating Signs

Problem: Find the nth term of the sequence: 1, -2, 3, -4, 5, ...

Solution Procedure:

1. Analyze the pattern:
   • Terms: 1, -2, 3, -4, 5.
   • Absolute values: 1, 2, 3, 4, 5 (increase by 1 each time, suggesting

     |a_n| = n

     ).
   • Signs: +, -, +, -, + (alternate, positive for odd ( n ), negative for even ( n )).
2. Formulate the nth term:
   • The magnitude is ( n ).
   • The sign is determined by

     (-1)^{n+1}

     : for

     n = 1

     ,

     (-1)^2 = 1

     ; for

     n = 2

     ,

     (-1)^3 = -1

     , etc.
   • Thus,

     a_n = (-1)^{n+1} \cdot n

     .
3. Verify:
   • n = 1

     :

     (-1)^2 \cdot 1 = 1

     .
   • n = 2

     :

     (-1)^3 \cdot 2 = -2

     .
   • n = 3

     :

     (-1)^4 \cdot 3 = 3

     .
   • n = 4

     :

     (-1)^5 \cdot 4 = -4

     .
   • n = 5

     :

     (-1)^6 \cdot 5 = 5

     .
4. State the answer: The nth term is

   (-1)^{n+1} \cdot n

   .

Detailed Explanation:

• Pattern recognition: Separating magnitude and sign simplifies the formula.
• Alternating signs: The term

  (-1)^{n+1}

  alternates between 1 and -1 based on ( n ).
• General form: Combines a linear term (( n )) with a sign factor.
• Formula:

  a_n = (-1)^{n+1} \cdot n

  .

Answer:

(-1)^{n+1} \cdot n

.

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Summary of Key Concepts

• Arithmetic sequences: Use

  a_n = a_1 + (n-1)d

  for constant differences.
• Geometric sequences: Use

  a_n = a_1 \cdot r^{n-1}

  for constant ratios.
• Quadratic sequences: Use

  a_n = an^2 + bn + c

  , identified by constant second differences.
• Recursive sequences: Compute terms iteratively using a recurrence relation.
• Pattern-based sequences: Derive formulas by analyzing magnitude, signs, or other patterns.
• System of equations: Solve for unknowns when terms are given but

  a_1

  , ( d ), or ( r ) are unknown.
• Verification: Always check the formula against known terms.

These problems cover a variety of sequence types and methods for finding the nth term, with detailed explanations to ensure clarity.