Question 3 – All Parts
Part (i)
📋 Given AP Sequence
\(3, 1, -1, -3, \ldots\)
🎯 To Find
- First term \((a)\)
- Common difference \((d)\)
💡 Concept Used
In an Arithmetic Progression:
- First term \((a)\) is the first number in the sequence
- Common difference \((d)\) is found by subtracting any term from the next term
📝 Step-by-Step Solution
The first term is the first number in the given sequence.
\[ a = 3 \]
Subtract the first term from the second term:
\[ \begin{aligned} d &= a_2 – a_1 \\ &= 1 – 3 \\ &= -2 \end{aligned} \]Check that the difference is constant for all consecutive terms:
\[ \begin{aligned} a_2 – a_1 &= 1 – 3 = -2 \quad ✓ \\ a_3 – a_2 &= -1 – 1 = -2 \quad ✓ \\ a_4 – a_3 &= -3 – (-1) = -3 + 1 = -2 \quad ✓ \end{aligned} \]Since the difference is constant (\(-2\)), this confirms it’s an AP.
We can also verify by checking if each term follows the formula \(a_n = a + (n-1)d\):
- \(a_1 = 3 + (1-1)(-2) = 3 + 0 = 3\) ✓
- \(a_2 = 3 + (2-1)(-2) = 3 – 2 = 1\) ✓
- \(a_3 = 3 + (3-1)(-2) = 3 – 4 = -1\) ✓
- \(a_4 = 3 + (4-1)(-2) = 3 – 6 = -3\) ✓
First term: \(a = 3\) | Common difference: \(d = -2\)
Part (ii)
📋 Given AP Sequence
\(-5, -1, 3, 7, \ldots\)
🎯 To Find
- First term \((a)\)
- Common difference \((d)\)
📝 Step-by-Step Solution
The first term is the first number in the sequence.
\[ a = -5 \]Subtract the first term from the second term:
\[ \begin{aligned} d &= a_2 – a_1 \\ &= -1 – (-5) \\ &= -1 + 5 \\ &= 4 \end{aligned} \]Check that the difference is constant for all consecutive terms:
\[ \begin{aligned} a_2 – a_1 &= -1 – (-5) = -1 + 5 = 4 \quad ✓ \\ a_3 – a_2 &= 3 – (-1) = 3 + 1 = 4 \quad ✓ \\ a_4 – a_3 &= 7 – 3 = 4 \quad ✓ \end{aligned} \]The common difference is constant (\(4\)), confirming this is an AP.
Notice that this AP is increasing because the common difference is positive (\(d = 4\)).
Each term is obtained by adding 4 to the previous term:
- \(-5 + 4 = -1\)
- \(-1 + 4 = 3\)
- \(3 + 4 = 7\)
- Next term: \(7 + 4 = 11\)
First term: \(a = -5\) | Common difference: \(d = 4\)
Part (iii)
📋 Given AP Sequence
\(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
🎯 To Find
- First term \((a)\)
- Common difference \((d)\)
💡 Special Note
This AP involves fractions. Since all terms have the same denominator (3), we can work with them directly.
📝 Step-by-Step Solution
The first term is:
\[ a = \frac{1}{3} \]Subtract the first term from the second term:
\[ \begin{aligned} d &= a_2 – a_1 \\ &= \frac{5}{3} – \frac{1}{3} \\ &= \frac{5 – 1}{3} \\ &= \frac{4}{3} \end{aligned} \]Check that the difference is constant:
\[ \begin{aligned} a_2 – a_1 &= \frac{5}{3} – \frac{1}{3} = \frac{4}{3} \quad ✓ \\ a_3 – a_2 &= \frac{9}{3} – \frac{5}{3} = \frac{4}{3} \quad ✓ \\ a_4 – a_3 &= \frac{13}{3} – \frac{9}{3} = \frac{4}{3} \quad ✓ \end{aligned} \]We can also simplify some terms to see the pattern more clearly:
- \(\frac{1}{3} = 0.333…\)
- \(\frac{5}{3} = 1.666…\)
- \(\frac{9}{3} = 3\)
- \(\frac{13}{3} = 4.333…\)
Common difference: \(\frac{4}{3} \approx 1.333…\)
First term: \(a = \frac{1}{3}\) | Common difference: \(d = \frac{4}{3}\)
Part (iv)
📋 Given AP Sequence
\(0.6, 1.7, 2.8, 3.9, \ldots\)
🎯 To Find
- First term \((a)\)
- Common difference \((d)\)
💡 Special Note
This AP involves decimal numbers. We can work with decimals directly for easier calculation.
📝 Step-by-Step Solution
The first term is:
\[ a = 0.6 \]Subtract the first term from the second term:
\[ \begin{aligned} d &= a_2 – a_1 \\ &= 1.7 – 0.6 \\ &= 1.1 \end{aligned} \]Check that the difference is constant for all consecutive terms:
\[ \begin{aligned} a_2 – a_1 &= 1.7 – 0.6 = 1.1 \quad ✓ \\ a_3 – a_2 &= 2.8 – 1.7 = 1.1 \quad ✓ \\ a_4 – a_3 &= 3.9 – 2.8 = 1.1 \quad ✓ \end{aligned} \]The common difference is constant (\(1.1\)).
We can also express these decimal values as fractions:
- \(0.6 = \frac{6}{10} = \frac{3}{5}\)
- \(1.7 = \frac{17}{10}\)
- \(2.8 = \frac{28}{10} = \frac{14}{5}\)
- \(3.9 = \frac{39}{10}\)
Common difference: \(1.1 = \frac{11}{10}\)
Next term prediction: \(3.9 + 1.1 = 5.0\)
First term: \(a = 0.6\) | Common difference: \(d = 1.1\)
⚠️ Common Mistakes to Avoid
❌ Mistake 1: Confusing the first term with the common difference.
✅ Correct: The first term \((a)\) is simply the first number in the sequence. The common difference \((d)\) is found by subtracting consecutive terms.
❌ Mistake 2: Sign errors when subtracting negative numbers.
✅ Correct: Remember: \(a – (-b) = a + b\). For example: \(-1 – (-5) = -1 + 5 = 4\).
❌ Mistake 3: Not verifying the common difference for all consecutive terms.
✅ Correct: Always check at least 2-3 consecutive differences to ensure they’re constant.
❌ Mistake 4: Calculation errors with fractions having the same denominator.
✅ Correct: When denominators are same: \(\frac{a}{c} – \frac{b}{c} = \frac{a-b}{c}\)
💡 Key Points to Remember
- The first term \((a)\) is always the first number in the sequence
- The common difference \((d)\) is calculated as: \(d = a_{n+1} – a_n\)
- The common difference must be constant for all consecutive terms in an AP
- Common difference can be positive, negative, or zero
- Common difference can be an integer, fraction, or decimal
- When \(d > 0\): AP is increasing
- When \(d < 0\): AP is decreasing
- When \(d = 0\): AP is a constant sequence
- Always verify by checking multiple consecutive differences
- For fractions with same denominator: subtract numerators directly
