Class 10 Maths Chapter 5 Exercise 5.2 Question 1 – Fill in the Blanks

Exercise 5.2 – Question 1

Fill in the blanks in the following table, given that \(a\) is the first term, \(d\) the common difference and \(a_n\) the \(n^{th}\) term of the AP.

Part (i)

Given: \(a = 7\), \(d = 3\), \(n = 8\)
Find: \(a_n = ?\)

📋 Given Information

  • First term: \(a = 7\)
  • Common difference: \(d = 3\)
  • Term number: \(n = 8\)

🎯 To Find

The 8th term of the arithmetic progression: \(a_8\)

📐 Formula Used

\[ a_n = a + (n – 1)d \]

where \(a_n\) is the \(n^{th}\) term, \(a\) is the first term, \(d\) is the common difference, and \(n\) is the term number.

📝 Step-by-Step Solution

1 Write the formula: \[ a_n = a + (n – 1)d \]
2 Substitute the given values: \[ \begin{aligned} a_8 &= 7 + (8 – 1) \times 3 \\ &= 7 + 7 \times 3 \\ &= 7 + 21 \\ &= 28 \end{aligned} \]
3 Verification:

Let’s verify by writing the first 8 terms:

\(7, 10, 13, 16, 19, 22, 25, 28\)

The 8th term is indeed 28. ✓

Answer: \(a_8 = 28\)

Part (ii)

Given: \(a = -18\), \(n = 10\), \(a_n = 0\)
Find: \(d = ?\)

📋 Given Information

  • First term: \(a = -18\)
  • Term number: \(n = 10\)
  • 10th term: \(a_{10} = 0\)

🎯 To Find

The common difference: \(d\)

📝 Step-by-Step Solution

1 Write the formula: \[ a_n = a + (n – 1)d \]
2 Substitute the known values: \[ \begin{aligned} 0 &= -18 + (10 – 1)d \\ 0 &= -18 + 9d \end{aligned} \]
3 Solve for \(d\): \[ \begin{aligned} 9d &= 18 \\ d &= \frac{18}{9} \\ d &= 2 \end{aligned} \]
4 Verification:

Let’s verify: \(a_{10} = -18 + (10-1) \times 2 = -18 + 18 = 0\) ✓

The AP is: \(-18, -16, -14, -12, -10, -8, -6, -4, -2, 0\)

Answer: \(d = 2\)

Part (iii)

Given: \(d = -3\), \(n = 18\), \(a_n = -5\)
Find: \(a = ?\)

📋 Given Information

  • Common difference: \(d = -3\)
  • Term number: \(n = 18\)
  • 18th term: \(a_{18} = -5\)

🎯 To Find

The first term: \(a\)

📝 Step-by-Step Solution

1 Write the formula: \[ a_n = a + (n – 1)d \]
2 Substitute the known values: \[ \begin{aligned} -5 &= a + (18 – 1)(-3) \\ -5 &= a + 17 \times (-3) \\ -5 &= a – 51 \end{aligned} \]
3 Solve for \(a\): \[ \begin{aligned} a &= -5 + 51 \\ a &= 46 \end{aligned} \]
4 Verification:

Let’s verify: \(a_{18} = 46 + (18-1) \times (-3) = 46 – 51 = -5\) ✓

The first few terms are: \(46, 43, 40, 37, 34, \ldots\)

Answer: \(a = 46\)

Part (iv)

Given: \(a = -18.9\), \(d = 2.5\), \(a_n = 3.6\)
Find: \(n = ?\)

📋 Given Information

  • First term: \(a = -18.9\)
  • Common difference: \(d = 2.5\)
  • nth term: \(a_n = 3.6\)

🎯 To Find

The term number: \(n\)

📝 Step-by-Step Solution

1 Write the formula: \[ a_n = a + (n – 1)d \]
2 Substitute the known values: \[ \begin{aligned} 3.6 &= -18.9 + (n – 1) \times 2.5 \\ 3.6 &= -18.9 + 2.5(n – 1) \end{aligned} \]
3 Solve for \(n\): \[ \begin{aligned} 3.6 + 18.9 &= 2.5(n – 1) \\ 22.5 &= 2.5(n – 1) \\ \frac{22.5}{2.5} &= n – 1 \\ 9 &= n – 1 \\ n &= 10 \end{aligned} \]
4 Verification:

Let’s verify: \(a_{10} = -18.9 + (10-1) \times 2.5 = -18.9 + 22.5 = 3.6\) ✓

Answer: \(n = 10\)

📊 Summary Table

Parta (First Term)d (Common Difference)n (Term Number)aₙ (nth Term)
(i)73828
(ii)-182100
(iii)46-318-5
(iv)-18.92.5103.6

⚠️ Common Mistakes to Avoid

❌ Mistake 1: Using \(n\) instead of \((n-1)\) in the formula.

✅ Correct: Always use \(a_n = a + (n-1)d\), not \(a_n = a + nd\).

❌ Mistake 2: Sign errors when dealing with negative values.

✅ Correct: Be careful with signs: \(-18 + 18 = 0\), not \(-18 – 18\).

❌ Mistake 3: Calculation errors with decimals.

✅ Correct: Double-check decimal arithmetic: \(22.5 ÷ 2.5 = 9\).

❌ Mistake 4: Forgetting to add 1 when finding \(n\).

✅ Correct: If \(n – 1 = 9\), then \(n = 10\), not 9.

💡 Key Points to Remember

  • The nth term formula: \(a_n = a + (n-1)d\)
  • To find \(a_n\): Substitute \(a\), \(d\), and \(n\) into the formula
  • To find \(d\): Rearrange to \(d = \frac{a_n – a}{n – 1}\)
  • To find \(a\): Rearrange to \(a = a_n – (n-1)d\)
  • To find \(n\): Rearrange to \(n = \frac{a_n – a}{d} + 1\)
  • Always verify your answer by substituting back into the formula
  • Be careful with negative numbers and decimal calculations

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