If the 3rd and 9th terms of an AP are 4 and – 8 respectively,

Solution:

Given that,

$a_{2}=4$

$a_{9}=-8$

We know that,

$a_{n}=a+(n-1) d$

$a_{2}=a+(3-1) d$

$4=a+2 d$ ...(i)

$a_{9}=a+(9-1) d$

$-8=a+8 d$ ...(ii)

On subtracting equation (I) from (II), we obtain

$-12=6 d$

$d=-2$

From equation (I), we obtain

$4=a+2(-2)$

$4=a+2(-2)$

$4=a-4$

$a=8$

Let $\mathrm{n}^{\text {th }}$ term of this A.P. be zero.

$a_{n}=a+(n-1) d$

0 = 8 + (n – 1) (–2)

0 = 8 – 2n + 2

2n = 10

n = 5

Hence, $5^{\text {th }}$ term of this A.P. is 0 .