Question

Find:
(i) 10th term of the A.P. 1, 4, 7, 10, ...
(ii) 18th term of the A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},$ ...
(iii) nth term of the A.P. 13, 8, 3, −2, ...

Answer 1

(i) 1, 4, 7, 10...

We have:
a = 1
d = $4-1=3$

$$\begin{gathered} a_{10}=a+(10-1)d\left[a_n=a+\left(n-1\right)d\right] \\ =a+9d \\ =1+9\times 3 \\ =28 \end{gathered}$$

(ii) $\sqrt{2},3\sqrt{2},5\sqrt{2}$...
We have:
$$\begin{gathered} a=\sqrt{2} \\ d=3\sqrt{2}-\sqrt{2}=2\sqrt{2} \end{gathered}$$

$$\begin{gathered} a_{18}=a+\left(18-1\right)d\left[a_n=a+\left(n-1\right)d\right] \\ =a+17d \\ =\sqrt{2}+17\left(2\sqrt{2}\right) \\ =\sqrt{2}+34\sqrt{2} \\ =35\sqrt{2} \end{gathered}$$

(iii) 13, 8, 3, −2...
We have:
$$\begin{gathered} a=13 \\ d=8-13=-5 \end{gathered}$$

$$\begin{gathered} a_n=a+(n-1)d \\ =13+(n-1)\left(-5\right) \\ =13-5n+5 \\ =18-5n \end{gathered}$$