Free Arithmetic Progressions 03 Practice Test - 10th Grade 

Let the numbers be $a-3d,a-d,a+d$ and $a+3d.$ These numbers form an AP with first term $a-3d$ and common difference $2d>0.$
Also, given that  $(a-3d)+(a-d)+(a+d)+(a+3d)=20.$
$\Rightarrow 4a=20$
$\Rightarrow a=5$
Also, ${(a-3d)}^2+{(a-d)}^2+{(a+d)}^2+{(a+3d)}^2=120$
$\Rightarrow 4a^2+20d^2=120$
$\Rightarrow 4(5{)}^2+20d^2=120$
$\Rightarrow d^2=1$ 
$\Rightarrow d=\pm 1$
SIince $d>0,$ we must have $d=1.$
Hence, the numbers are 2, 4, 6, and 8.

$34+32+30+....+10$
This sequence is an AP.
Here, $a=34,d=32-34=-2.$
Let the number of terms of the AP be $n.$
We know that
$a_n=a+(n-1)d.$

$\Rightarrow 10=34+(n-1)(-2)$

$\Rightarrow (n-1)(-2)=-24$

$\Rightarrow n-1=\frac{24}{2}=12$

$\Rightarrow n=13$ 

Also, we know that the sum to $n$ terms of an AP with first term $a$ and last term $l$ is given by

     $S_n=\frac{n}{2}(a+l).$

$\therefore S_{13}=\frac{13}{2}(34+10)$

    $S_{13}=286$

Hence, the required sum is 286. 
 

The series can be clubbed into two AP's, (1 + 7 + 13......) and (4 + 5 + 6.......). So, we can observe that the series can be taken as two AP's by simple rearrangement of numbers. Now, we can find the Sum of 50 terms for each AP formed.

Formula for sum of first n terms of AP is:$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

So, for first AP
$S_1=\frac{50}{2}(2+49\times 6)=7400$
and similarly,
$S_2=\frac{50}{2}(8+49)=1425$

So, Sum of first 100 terms
$=7400+1425=8825$

Let $S_{10}$ be the sum of first 10 terms and $S_5$ be the sum of first 5 terms.
The sum upto $n$ terms of an AP of first term $a$ and common difference $d$ is given by
$S_n=\frac{n}{2}[2a+(n-1\left)d\right].$

Given, $S_{10}=4S_5.$

$\Rightarrow \frac{10}{2}[2a+(10-1\left)d\right]=4\times \frac{5}{2}[2a+(5-1\left)d\right]$

$\Rightarrow \frac{10}{2}[2a+9d]=4\times \frac{5}{2}[2a+4d]$
$\Rightarrow 2a+9d=4a+8d$
$\Rightarrow 2a=d$
$\Rightarrow \frac{a}{d}=\frac{1}{2}$

It is also given that $a_3=12$  and $a_{50}=106$      
Using formula $a_n=a+(n-1)d$ to find $n^{th}$ term of AP, we get
$a_{50}=a+(50-1)d\Rightarrow 106=a+49d\cdots \left(1\right)$  
$a_3=a+(3-1)d\Rightarrow 12=a+2d\cdots \left(2\right)$

Subtracting (1) and (2), we get,
$\Rightarrow 47d=94$
$\Rightarrow d=2$

Substituting $d$ in (2), we get,
$a+2\left(2\right)=12$
$a=12-4=8$

The ${29}^{th}$ term is,
$a_{29}=a+(29-1)d=8+28\left(2\right)=8+56=64$

It is given that ${17}^{th}$ term exceeds its ${10}^{th}$ term by 7
It means $a_{17}$ = $a_{10}$ + 7......(1)
Here, $a_{17}$ is the ${17}^{th}$ term and a₁₀ is the ${10}^{th}$ term of an AP.
Using formula $a_n=a+(n-1)d$   to find $n^{th}$ term of arithmetic progression, we get

$a_{17}$ =$a+\left(16\right)d$  .....(2)

$a_{10}$ = $a+\left(9\right)d$ .....(3)

Putting (2) and (3) in equation (1), we get

$a+16d=a+9d+7$

$\Rightarrow 7d=7$

$\Rightarrow d=\frac{7}{7}=1$

There are five terms in this AP
$a_n$ = $a_1$ + (n - 1)d
$\Rightarrow$ $a_n-a_1=(n-1)d$
$\Rightarrow$  $\frac{a_n-a_1}{n-1}=d$

Here, $a_1=1;n=5;a_n=9$
Therefore the common difference,
$d=\frac{9-1}{5-1}=2$