Question

If the sixth term of an AP is zero then show that its 33rd term is threе times its 15th term.

Answer 1

Given: Let the AP have first term a and common difference d. The nth term is T_n = a + (n – 1)d. It is given that the 6th term is zero, T₆ = 0.

Step-wise calculation:

1. T₆ = a + 5d = 0

⇒ a = –5d

2. T₁₅ = a + 14d

= (–5d) + 14d

= 9d

3. T₃₃ = a + 32d

= (–5d) + 32d

= 27d

4. Hence, T₃₃ = 27d

= 3 × 9d

= 3 × T₁₅

Therefore, the 33rd term is three times the 15th term.