Solution:
given , 3rd term of an AP is 4 and 9th term of an AP is -8Nth term of an AP is an
an = a+ (n-1)d
Where a is the first term
Here 3rd term of an AP = 4 n= 3
I.e a3= a(3-1)d=4
a3= a+2d= 4
a+2d = 4..(1)
9th term of an AP = -8
n=9
a9= a(9-1)d= -8
a9= a+8d= -8
a+8d = -8..(2)
Now we will solve equation (1)and(2)
(a+2d)-(a+8d)=4-(-8)
a+2d-a-8d= 12
-6d=12
d=-2
Substituting the value of d in equation..(1)
a+2d= 4
a+2(-2)=4
a-4=4
a=8
Now, we will substitute the value of a and d
a+(n-1)d=0
8+(n-1)(-2)=0
8-2n+2=0
10-2n=0
10=2n
10/2=n
∴ therefore, the 5th term of an AP is zero.
