NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.1
Question 6

Question.  6

The 21st term of the AP whose first two terms are \(-3\) and \(4\) is

(A)

17

(B)

137

(C)

143

(D)

−143

Detailed Answer with Explanation:

Step 1: In an Arithmetic Progression (AP), the first term is called \(a\). Here, the first term is \(-3\). So, \(a = -3\).

Step 2: The common difference \(d\) is found by subtracting the first term from the second term.
\(d = 4 - (-3) = 4 + 3 = 7\).

Step 3: The formula for the \(n\)-th term of an AP is:
\(a_n = a + (n-1)\,d\).

Step 4: We need the 21st term. So, put \(n = 21\).
\(a_{21} = a + (21 - 1) d\).

Step 5: Substitute the values:
\(a_{21} = -3 + 20 \times 7\).

Step 6: Multiply: \(20 \times 7 = 140\).

Step 7: Add: \(-3 + 140 = 137\).

Final Answer: The 21st term is 137.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 5: Arithematic Progressions – Exercise 5.1 | Detailed Answers