Question

(i) Is 68 a term of the A.P. 7, 10, 13, ...?
(ii) Is 302 a term of the A.P. 3, 8, 13, ...?

Answer 1

(i) 7, 10, 13...
Here, we have:
a = 7
$$\begin{gathered} d=\left(10-7\right)=3 \\ \mathrm{Let}{a}_n=68 \\ \Rightarrow a+(n-1)d=68 \\ \Rightarrow 7+(n-1)\left(3\right)=68 \\ \Rightarrow (n-1)\left(3\right)=61 \\ \Rightarrow (n-1)=\frac{61}{3} \\ \Rightarrow n=\frac{61}{3}+1=\frac{64}{3} \end{gathered}$$

Since n is not a natural number.So, 68 is not a term of the given A.P.

(ii) 3, 8, 13...
Here, we have:
a = 3
$$\begin{gathered} d=\left(8-3\right)=5 \\ \mathrm{Let}{a}_n=302 \\ \Rightarrow a+\left(n-1\right)d=302 \\ \Rightarrow 3+\left(n-1\right)5=302 \\ \Rightarrow \left(n-1\right)5=299 \\ \Rightarrow \left(n-1\right)=\frac{299}{5} \\ \Rightarrow n=\frac{299}{5}+1=\frac{304}{5} \end{gathered}$$
Since n is not a natural number.So, 302 is not a term of the given A.P.