Find:

In this problem, we are given different A.P. and we need to find the required term of that A.P.

(i) $10^{\text {th }}$ term of the A.P. $1,4,7,10, \ldots$

Here,

First term (a) = 1

Common difference of the A.P. $(d)=4-1$

$=3$

Now, as we know,

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

So, for $10^{\text {th }}$ term,

$a_{10}=a+(10-1) d$

$=1+(9)^{3}$

$=1+27$

$=28$

Therefore, the $10^{\text {th }}$ term of the given A.P. is $a_{10}=28$.

(ii) $18^{\text {th }}$ term of the A.P. $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$

Here,

First term $(a)=\sqrt{2}$

Common difference of the A.P. $(d)=3 \sqrt{2}-\sqrt{2}$

$=2 \sqrt{2}$

Now, as we know,

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

So, for $18^{\text {th }}$ term,

$a_{18}=a+(18-1) d$

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

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

$=35 \sqrt{2}$

Therefore, the $18^{\text {th }}$ term of the given A.P. is $a_{18}=35 \sqrt{2}$.

(iii) $n^{\text {th }}$ term of the A.P. $13,8,3,-2, \ldots$

Here,

First term (a) = 13

Common difference of the A.P. (d) 

Now, as we know,

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

So, for $n^{\text {th }}$ term,

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

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

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

$=13-5 n+5$

$=18-5 n$

Therefore, the $n^{\text {th }}$ term of the given A.P. is $a_{n}=18-5 n$.

(iv) $10^{\text {th }}$ term of the A.P. $-40,-15,10,35, \ldots$

Here,

First term $(a)=-40$

Common difference of the A.P. $(d)=-15-(-40)$

$=-15+40$

$=25$

Now, as we know,

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

So, for $10^{\text {th }}$ term,

$a_{10}=a+(10-1) d$

$=-40+(9) 25$

$=-40+225$

$=185$

Therefore, the $10^{\text {th }}$ term of the given A.P. is $a_{10}=185$.

(v) $8^{\text {th }}$ term of the A.P. $117,104,91,78 \ldots$

Here,

First term (a) = 117

Common difference of the A.P. $(d)=104-117$

$=-13$

Now, as we know,

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

So, for $8^{\text {th }}$ term,

$a_{8}=a+(8-1) d$

$=117+(7)(-13)$

$=117-91$

$=26$

Therefore, the $8^{\text {th }}$ term of the given A.P. is $a_{x}=26$.

(vi) $11^{\text {th }}$ term of the A.P. $10.0,10.5,11.0,11.5, \ldots$

Here,

First term (a) = 10.0

Common difference of the A.P. $(d)=10.5-10.0$

$=0.5$

Now, as we know,

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

So, for $11^{\text {th }}$ term,

$a_{11}=a+(11-1) d$

$=10.0+(10)(0.5)$

$=10.0-5.0$

$=15.0$

Therefore, the $11^{\text {th }}$ term of the given A.P. is $a_{11}=15.0$.

(vii) $9^{\text {th }}$ term of the A.P. $\frac{3}{4}, \frac{5}{4}, \frac{7}{4}, \frac{9}{4}, \ldots$

Here,

First term $(a)=\frac{3}{4}$

Common difference of the A.P. $(d)=\frac{5}{4}-\frac{3}{4}$

$=\frac{5-3}{4}$

$=\frac{2}{4}$

Now, as we know,

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

So, for $9^{\text {th }}$ term,

$a_{9}=a+(9-1) d$

$=\frac{3}{4}+(8)\left(\frac{2}{4}\right)$

$=\frac{3}{4}+\frac{16}{4}$

$=\frac{19}{4}$

Therefore, the $9^{\text {th }}$ term of the given A.P. is $a_{9}=\frac{19}{4}$.