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drills arithmetic sequences & series 1. the 12th term of \\(7,11,15,19,…

Question

drills

arithmetic sequences & series

  1. the 12th term of \\(7,11,15,19,\dots\\) is: a) 51 b) 55 c) 59 d) 63
  2. the sum of the first 25 terms of \\(4,9,14,19,\dots\\) is: a) 1500 b) 1600 c) 1700 d) 1800
  3. the common difference of \\(30,25,20,15,\dots\\) is: a) -10 b) -5 c) 5 d) 10
  4. the sum of the first 20 terms of \\(3,7,11,15,\dots\\) is: a) 800 b) 810 c) 820 d) 830
  5. a student saves \\(\text{p}50\\) in week 1, \\(\text{p}75\\) in week 2, \\(\text{p}100\\) in week 3, and continues increasing by \\(\text{p}25\\) weekly. how much in 12 weeks? a) 2000 b) 2100 c) 2200 d) 2250

Explanation:

🆕 New Concept Discovered: Arithmetic Sequences and Series
Finding terms and sums of sequences with a constant difference

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Step 1: Solve Question 1

To find the \(n\)-th term of an arithmetic sequence, we use the formula:

$$ a_n = a_1 + (n - 1)d $$

where \(a_1\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

  • Sequence: \(7, 11, 15, 19, \dots\)
  • First term (\(a_1\)): \(7\)
  • Common difference (\(d\)): \(11 - 7 = 4\)
  • Term to find (\(n\)): \(12\)

Substitute the values into the formula:

$$ a_{12} = 7 + (12 - 1) \cdot 4 $$
$$ a_{12} = 7 + 11 \cdot 4 $$
$$ a_{12} = 7 + 44 = 51 $$

This matches option A.

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Step 2: Solve Question 2

To find the sum of the first \(n\) terms of an arithmetic sequence, we use the sum formula:

$$ S_n = \frac{n}{2} [2a_1 + (n - 1)d] $$
  • Sequence: \(4, 9, 14, 19, \dots\)
  • First term (\(a_1\)): \(4\)
  • Common difference (\(d\)): \(9 - 4 = 5\)
  • Number of terms (\(n\)): \(25\)

Substitute the values into the formula:

$$ S_{25} = \frac{25}{2} [2(4) + (25 - 1) \cdot 5] $$
$$ S_{25} = 12.5 [8 + 24 \cdot 5] $$
$$ S_{25} = 12.5 [8 + 120] $$
$$ S_{25} = 12.5 \cdot 128 = 1600 $$

This matches option B.

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Step 3: Solve Question 3

The common difference (\(d\)) is found by subtracting any term from the term that immediately follows it:

$$ d = a_{n} - a_{n-1} $$
  • Sequence: \(30, 25, 20, 15, \dots\)
  • Calculation:
$$ d = 25 - 30 = -5 $$

This matches option B.

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Step 4: Solve Question 4

We use the sum formula again:

$$ S_n = \frac{n}{2} [2a_1 + (n - 1)d] $$
  • Sequence: \(3, 7, 11, 15, \dots\)
  • First term (\(a_1\)): \(3\)
  • Common difference (\(d\)): \(7 - 3 = 4\)
  • Number of terms (\(n\)): \(20\)

Substitute the values:

$$ S_{20} = \frac{20}{2} [2(3) + (20 - 1) \cdot 4] $$
$$ S_{20} = 10 [6 + 19 \cdot 4] $$
$$ S_{20} = 10 [6 + 76] $$
$$ S_{20} = 10 \cdot 82 = 820 $$

This matches option C.

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Step 5: Solve Question 5

We need to find the total amount saved over \(12\) weeks. This is the sum of an arithmetic series where each week's savings is a term.

  • Savings sequence: \(50, 75, 100, \dots\)
  • First term (\(a_1\)): \(50\)
  • Common difference (\(d\)): \(25\)
  • Number of terms (\(n\)): \(12\)

Substitute the values into the sum formula:

$$ S_{12} = \frac{12}{2} [2(50) + (12 - 1) \cdot 25] $$
$$ S_{12} = 6 [100 + 11 \cdot 25] $$
$$ S_{12} = 6 [100 + 275] $$
$$ S_{12} = 6 \cdot 375 = 2250 $$

This matches option D.

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Answer:

  1. A) 51
  2. B) 1600
  3. B) -5
  4. C) 820
  5. D) 2250