NCERT Solutions
Class 10 - Mathematics - Chapter 5: ARITHMETIC PROGRESSIONS - Exercise 5.2

Question 5

Recall: For an AP with first term \(a\), common difference \(d\), and \(n\) terms, the \(n\)th term is given by

\[a_n = a + (n - 1)d\]

When the last term is known, we set \(a_n\) equal to that last term and solve for \(n\).

(i) AP: 7, 13, 19, ..., 205

Step 1: Identify \(a\), \(d\), and \(a_n\).

First term: \(a = 7\).

Common difference: \(d = 13 - 7 = 6\).

Last term: \(a_n = 205\).

Step 2: Use the nth-term formula.

\[a_n = a + (n - 1)d\]

Substitute the values:

\[205 = 7 + (n - 1) \cdot 6\]

Step 3: Solve for \(n\).

Subtract 7 from both sides:

\[205 - 7 = 6(n - 1)\]

\[198 = 6(n - 1)\]

Divide by 6:

\[n - 1 = \dfrac{198}{6} = 33\]

So,

\[n = 33 + 1 = 34\]

Conclusion for (i): The AP has 34 terms.

(ii) AP: 18, 15 1/2, 13, ..., -47

Write the mixed fraction \(15\tfrac{1}{2}\) as an improper fraction: \(15\tfrac{1}{2} = \dfrac{31}{2}\).

So the AP is: \(18, \dfrac{31}{2}, 13, \ldots, -47\).

Step 1: Identify \(a\), \(d\), and \(a_n\).

First term: \(a = 18\).

Common difference:

\[d = \dfrac{31}{2} - 18 = \dfrac{31}{2} - \dfrac{36}{2} = -\dfrac{5}{2}\]

Last term: \(a_n = -47\).

Step 2: Use the nth-term formula.

\[a_n = a + (n - 1)d\]

Substitute the values:

\[-47 = 18 + (n - 1)\left(-\dfrac{5}{2}\right)\]

Step 3: Solve for \(n\).

Subtract 18 from both sides:

\[-47 - 18 = (n - 1)\left(-\dfrac{5}{2}\right)\]

\[-65 = (n - 1)\left(-\dfrac{5}{2}\right)\]

Multiply both sides by 2 to clear the denominator:

\[-130 = (n - 1)(-5)\]

Divide both sides by -5:

\[n - 1 = \dfrac{-130}{-5} = 26\]

So,

\[n = 26 + 1 = 27\]

Conclusion for (ii): The AP has 27 terms.

Final Answer: (i) 34 terms, (ii) 27 terms.