NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 5: Arithematic Progressions - Exercise 5.1
Question 7

Question.  7

If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?

(A)

30

(B)

33

(C)

37

(D)

38

Detailed Answer with Explanation:

Step 1: Recall the formula for the nth term of an AP: \(a_n = a + (n-1) d\), where \(a\) is the first term, and \(d\) is the common difference.

Step 2: The 2nd term is given as 13. Using the formula: \(a_2 = a + (2-1)d = a + d\). So, \(a + d = 13\). (Equation 1)

Step 3: The 5th term is given as 25. Using the formula: \(a_5 = a + (5-1)d = a + 4d\). So, \(a + 4d = 25\). (Equation 2)

Step 4: Subtract Equation (1) from Equation (2): \((a + 4d) - (a + d) = 25 - 13\) \(3d = 12\) \(d = 4\).

Step 5: Put the value of \(d\) into Equation (1): \(a + 4 = 13\) \(a = 9\).

Step 6: Now find the 7th term: \(a_7 = a + (7-1)d = 9 + 6 \times 4 = 9 + 24 = 33\).

Final Answer: The 7th term is 33.

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