Free Arithmetic Progressions 02 Practice Test - 10th Grade 

 Since, 2x, x + 10, 3x + 2 are in AP.

2 (x + 10) = 2x + (3x + 2)

2x + 20 = 5x + 2

3x = 18

x = 6.

(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,
 
$AM=\frac{\text{Sum of the numbers}}{\text{Number of numbers}}$

$=\frac{a+b}{2}$