Question

(i) Which term of the A.P. 3, 8, 13, ... is 248?
(ii) Which term of the A.P. 84, 80, 76, ... is 0?
(iii) Which term of the A.P. 4, 9, 14, ... is 254?

Answer 1

(i) 3, 8, 13...
Here, we have:
a = 3
$d=\left(8-3\right)=5$
$$\begin{gathered} \mathrm{Let}{a}_n=248 \\ \Rightarrow a+\left(n-1\right)d=248 \\ \Rightarrow 3+\left(n-1\right)5=248 \\ \Rightarrow \left(n-1\right)5=245 \\ \Rightarrow n-1=49 \\ \Rightarrow n=50 \end{gathered}$$

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

(ii) 84, 80, 76...
Here, we have:
a = 84
$d=\left(80-84\right)=-4$
$$\begin{gathered} Let{a}_n=0 \\ \Rightarrow a+(n-1)d=0 \\ \Rightarrow 84+(n-1)\left(-4\right)=0 \\ \Rightarrow (n-1)\left(-4\right)=-84 \\ \Rightarrow (n-1)=21 \\ \Rightarrow n=22 \end{gathered}$$

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

(iii) 4, 9, 14...
Here, we have:
a = 4
$d=\left(9-4\right)=5$
$$\begin{gathered} Let{a}_n=254 \\ \Rightarrow a+\left(n-1\right)d=254 \\ \Rightarrow 4+\left(n-1\right)5=254 \\ \Rightarrow \left(n-1\right)5=250 \\ \Rightarrow \left(n-1\right)=50 \\ \Rightarrow n=51 \end{gathered}$$

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