Question

the 9th ye of an A.P is 499 and the 499th term is 9. which term of this A.P is 0​

Answer 1

Step-by-step explanation:

Let the first term of AP = a.

and the common difference = d.

given that a9=499

Here n value is 9. Put value in a+(n-1)d

a9 = a + 8d = 499

Therefore, a + 8d = 499 (1)

a499=9

a499 = a + 498d = 9.

Therefore, a + 498d = 9 (2)

Subtracting eq(1) from eq(2)

a+498d-a-8d=9-499

490d=−490d=−490490=−1

Therefore, common difference, d = -1.

Substituting the value of d in eq(1).

⇒a+8d=499⇒a+(8∗(−1))=499.⇒a=499+8⇒a=507

Therefore, first term, a = 507.

The required term = an

and an = 0

a+(n−1)d=0.

⇒ Putting value of a and d

⇒507+(n−1)−1=0

⇒507=n−1

⇒n=507+1