Find:

Solution:

(i) $1,4,7,10 \ldots$

We have;

$a=1$

$d=4-1=3$

$a_{10}=a+(10-1) d \quad\left[a_{n}=a+(n-1) d\right]$

$=a+9 d$

$=1+9 \times 3$

$=28$

(ii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2} \ldots$

We have;

$a=\sqrt{2}$

$d=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$

$a_{18}=a+(18-1) d \quad\left[a_{n}=a+(n-1) d\right]$

$=a+17 d$

$=\sqrt{2}+17(2 \sqrt{2})$

$=\sqrt{2}+34 \sqrt{2}$

$=35 \sqrt{2}$

(iii) 13, 8, 3, −2...

We have:

$a=13$

$d=8-13=-5$

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

$=13+(n-1)(-5)$

$=13-5 n+5$

$=18-5 n$