Question 42 of 44intermediate🔧 ApplyShort Answer3 marks

Which term of the AP: 121, 117, 113, …, is its first negative term?

Correct Answer

The 32nd term is the first negative term of the given AP.

Given AP: 121, 117, 113, …

Here, first term (a) = 121 and common difference (d) = 117 – 121 = –4.

Using the formula for the nth term: aₙ = a + (n – 1)d

For the first negative term, we need aₙ < 0

So, 121 + (n – 1)(–4) < 0

⇒ 121 – 4(n – 1) < 0

⇒ 121 – 4n + 4 < 0

⇒ 125 – 4n < 0

⇒ 125 < 4n

⇒ n > 31.25

Since n must be an integer, the smallest integer greater than 31.25 is n = 32.

Verification: a₃₂ = 121 + 31(–4) = 121 – 124 = –3 (negative)

And a₃₁ = 121 + 30(–4) = 121 – 120 = 1 (positive)

Therefore, the 32nd term is the first negative term.

Exercise: EXERCISE 5.4 (Optional) | Q: 1 | (Page Page 23)
For More Understanding

Explanation

This question is from Exercise 5.4 (Optional) of the NCERT textbook. The hint provided in the context guides students to find n for aₙ < 0. The solution follows the same method as Example 4, where we find which term equals a specific value, but here we find the smallest n where the term becomes negative. The key insight is that we need the smallest integer n satisfying the inequality aₙ < 0.

Solution Steps

  1. Step 1: Identify a = 121, d = 117 – 121 = –4

  2. Step 2: Set up inequality aₙ = a + (n – 1)d < 0

  3. Step 3: Substitute values: 121 + (n – 1)(–4) < 0

  4. Step 4: Solve: 125 – 4n < 0, giving n > 31.25

  5. Step 5: Conclude n = 32 is the smallest integer satisfying the condition

PreviousNext