(i)  $7,13,\mathrm{19....},205$

First term, $a=7$

Common difference, $d=13-7=6$

$a_n=205$

Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get

$205=7+(n-1)6$
$\Rightarrow 205=6n+1$

$\Rightarrow 204=6n$

$\Rightarrow n=\frac{204}{6}=34$

Therefore, there are 34 terms in the given arithmetic progression.

(ii) $18,\frac{31}{2},13,...-47$

First term, $a=18$

Common difference, $d=\frac{31}{2}-18=\frac{-5}{2}$

$a_n$ = − 47

Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of arithmetic progression, we get

$-47=18+(n-1)\left(\frac{-5}{2}\right)$

$\Rightarrow -94=36-5n+5$

$\Rightarrow 5n=135$
$\Rightarrow n=\frac{135}{5}=27$

Therefore, there are 27 terms in the given arithmetic progression.

Here, First term, $a=3$

Common difference, $d=8-3=5$
$a_n=78$
Using formula $a_n=a+(n-1)d$ to find nth term of arithmetic progression, we get
$a_n=3+(n-1)5$  {we want to find value of n here.}
$\Rightarrow 78=3+(n-1)5$
$\Rightarrow 75=5n-5$
$\Rightarrow 80=5n$
$\Rightarrow n=\frac{80}{5}=16$
It means ${16}^{th}$ term of the given AP is equal to 78.

Length of semi-circle =$\frac{circumferenceofcircle}{2}$ = $\frac{2\pi r}{2}$ =  $\pi r$ .

Length of semi-circle of radii 0.5 cm = $\pi \times 0.5cm$

Length of semi-circle of radii 1.0 cm = $\pi \times 1.0cm$

Length of semi-circle of radii 1.5 cm = $\pi \times 1.5cm$

Length of semi-circle of radii 2.0 cm = $\pi \times 2.0cm$

.....and so on

$\pi \left(0.5\right),\pi \left(1.0\right),\pi \left(1.5\right).....$   13 term (There are total of thirteen semi-circles}. 
For total length of the spiral, we need to find sum of the sequence   13 terms 
Total length of spiral = $0.5\pi +\pi +1.5\pi +2\pi ........upto13terms$

$\Rightarrow$ Total length of spiral = $\pi (0.5+1.0+1.5+.......)upto13terms$

Sequence 0.5, 1.0, 1.5 ....13 terms is an arithmetic progression.

Let's find the sum of this sequence. 

$a=0.5;d=0.5$

$S=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow S=\frac{13}{2}(2\times 0.5+(13-1\left)\right(0.5\left)\right)$

$\Rightarrow S=\frac{13}{2}(1+6)$

$\Rightarrow S=\frac{91}{2}=45.5$

So 

$\Rightarrow$ Total length of spiral = $\pi (0.5+1.0+1.5+.......)$

$\Rightarrow$ Total length of spiral = $\pi \left(45.5\right)=143cm$

The given sequence is  5, 2, -1, -4, -7, ...., an AP with first term $a=5$ and common difference $d=2-5=-3.$

The sum to $n$ terms of an AP of first term $a$ and common difference $d$ is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right].\Rightarrow S_n=\frac{n}{2}[10+(n-1\left)\right(-3\left)\right]$
$=\frac{n}{2}[10-3n+3]$
$=\frac{n}{2}[13-3n]$

As $a_n=3+2n$,

so,

$a_1=3+2\times 1=5$

$a_2=3+2\times 2=7$

$a_3=3+2\times 3=9$

List of numbers becomes 5, 7, 9, 11 . . .

Here, $7–5=9–7=11–9=2$ and so on.

So, it forms an AP with common difference $d=2$.

To find $S_{24}$, we have $n=24,a=5,d=2$.

Therefore, $S_{24}=\frac{24}{2}\left(2\right(5)+(24-1\left)2\right)$

                         $=12(10+46)=672$

The arithmetic mean of two numbers is the simple average of the two numbers.
Thus,
 
$AM=\frac{\text{Sum of the numbers}}{\text{Number of numbers}}$
$=\frac{a+b}{2}$

Given, $x+1,3x,4x+2$ are in AP.
So, the difference of any two consecutive terms will be the same.
$\therefore 3x–(x+1)=(4x+2)–3x$
$\Rightarrow 2x-1=x+2$
$\Rightarrow x=3$

So, the first term of the AP is $a=x+1=3+1=4.$
The second term of the AP is $3x=3\times 3=9.$
$\therefore \text{Common difference,}d=9–4=5$

The $n^{\text{th}}$ term of an AP with first term $a$ and common difference $d$ is given by
     $t_n=a+(n-1)d.$
$\Rightarrow t_5=4+4\left(5\right)=24$

Given that
$a_{11}=38$ and  $a_{16}=73,$
where $a_{11}$ is the ${11}^{\text{th}}$ term and $a_{16}$ is the ${16}^{\text{th}}$ term of an AP.
The $n^{\text{th}}$ term of an AP with first term $a$ and common difference $d$ is given by
$a_n=a+(n-1)d.$
$\Rightarrow 38=a+10d....\left(i\right)$
$73=a+15d....\left(ii\right)$
Subtracting equation (i) from equation (ii), we get
$35=5d.$
$\Rightarrow d=7$
Substituting the value of $d$ in equation (i), we get
$a=38-10d=38-10\left(7\right)=-32.$
Now, $a_{31}=a+(31-1)d=-32+30\left(7\right)$
       $=178$
Therefore, the ${31}^{\text{st}}$ term of the given AP is 178.

Given, $a_3=4$ and $a_9=-8,$ where $a_3$ and $a_9$ are the third and ninth terms of an AP respectively.
Using the formula for $n^{\text{th}}$ term of an AP with first term $a$ and common difference $d,$
$a_n=a+(n-1)d,$ we get
$a_3=a+(3-1)d$ and $a_9=a+(9-1)d.$
$\Rightarrow 4=a+2d...\left(i\right)$
$-8=a+8d...\left(ii\right)$
Substituting the value of $a$ from equation (i) in equation (ii), we have
        $-8=4-2d+8d$
$\Rightarrow -12=6d$
$\Rightarrow d=-\frac{12}{6}=-2$
Solving for $a,$ we get $-8=a-16.$
$\Rightarrow a=8$
Therefore, the first term of the AP is 8 and common difference is −2.
Let the $n^{\text{th}}$ term of the AP be zero.
i.e.,                  $a_n=0$ 
$a+(n-1)d=0$
$8+(n-1)(-2)=0$
$\Rightarrow 8-2n+2=0$
$\Rightarrow 2n=10$
$\Rightarrow n=\frac{10}{2}=5$
Therefore, the $5^{\text{th}}$ term of the AP is equal to 0.