Solve the following

(i) 3, 8, 13...

Here, we have:

a = 3

$d=(8-3)=5$

Let $a_{n}=248$

$\Rightarrow a+(n-1) d=248$

$\Rightarrow 3+(n-1) 5=248$

$\Rightarrow(n-1) 5=245$

$\Rightarrow n-1=49$

$\Rightarrow n=50$

Hence, 248 is the 50th term of the given A.P.

(ii) 84, 80, 76...

Here, we have:

a = 84

$d=(80-84)=-4$

Let $a_{n}=0$

$\Rightarrow a+(n-1) d=0$

$\Rightarrow 84+(n-1)(-4)=0$

$\Rightarrow(n-1)(-4)=-84$

$\Rightarrow(n-1)=21$

$\Rightarrow n=22$

Hence, 0 is the 22nd term of the given A.P.

(iii) 4, 9, 14...

Here, we have:

a = 4

$d=(9-4)=5$

Let $a_{n}=254$

$\Rightarrow a+(n-1) d=254$

$\Rightarrow 4+(n-1) 5=254$

$\Rightarrow(n-1) 5=250$

$\Rightarrow(n-1)=50$

$\Rightarrow n=51$

Hence, 254 is the 51st term of the given A.P.