Question

2. If the third and the ninth
terms of an AP are 4, and 8
respectively, which term of
this AP is zero?​

Answer 1

Answer:

Let first term be a and common difference be d

Then, a3= a+(3-1)d= a+2d = 4

and, a9 = a+(9-1)d = a+8d= 8

Now, a9-a3 = a+8d-(a+2d) = 4 which gives d = 2/3

So, putting d in a3 we get, a = 4-2.(2/3) = 8/3

Now, an = 0

So, 8/3 + (n-1)(2/3) = 0

=> 2n/3 + 2= 0

=> n = -3 which is not possible.

Hence, no term of the AP is zero.