Free Arithmetic Progressions 01 Practice Test - 10th Grade 

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.

Natural numbers that lie between 24 and 101,which are divisible by 5 are 25, 30,... and 100.
This is an A.P with first term $a=25,$ common difference $d=5$ and the last term $l=100.$
  But, $l=a+(n-1)d.$
$\Rightarrow 100=25+(n-1)\times 5$
$\Rightarrow 75=(n-1)\times 5$
$\Rightarrow 15=(n-1)$
$\Rightarrow 16=n$
Thus, there are 16 natural numbers between 24 and 101 that are divisible by 5.

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.

Here, $S_{14}$ = 1050, n = 14, a = 10.

As $S_n$ = ($\frac{n}{2}$) [2a + (n-1) d]

So,
$1050=(\frac{14}{2}$) [20 + 13d]

$\Rightarrow 1050=140+91d$

$\Rightarrow 910=91d$

$\Rightarrow d=10$

Therefore, $a_{20}=10+(20–1)\times 10=200$

i.e. ${20}^{th}$   term is 200.

Using the formula
$S_n=\frac{n}{2}[2n+(n-1\left)d\right]$
Sum of first 100 natural numbers = $\frac{100}{2}(2+99)=5050$

Sum of even numbers from 1 to 100, there are 50 even numbers between 1 and 100.

So, we have
Sum of even numbers from 1 to 100 = $\frac{50}{2}(2\times 2+49\times 2)$ = 2550

Hence, the difference is 5050 - 2550 = 2500

Given, $t_n=3+2n.$

$\therefore t_1=3+2\times 1=5$

    $t_2=3+2\times 2=7$

    $t_3=3+2\times 3=9$

Note that $t_2-t_1=t_3-t_2=2.$

Thus, the common difference of the AP is 2.

Therefore, the AP is 5, 7, 9, 11 . . .

The sum to $n$ terms of an AP with first term $a$ and common difference $d$ is

$S_n=\frac{n}{2}[2a+(n-1\left)d\right].$

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

           $=12(10+46)=672$

The first 15 multiples of 8 are 8, 16, 24, 32, .... 112, 120

First term, $a=8$

Common difference, $d=16-8=8$

$n=15$

Applying formula, $S_n=\frac{n}{2}(2a+(n-1\left)d\right)$ to find sum of $n$ terms of AP, we get
$S_{15}=\frac{15}{2}\left(2\right(8)+(15-1\left)\right(8\left)\right)$
       $=\frac{15}{2}(16+8(14\left)\right)$
       $=\frac{15}{2}(16+112)$
       $=\frac{15}{2}\left(128\right)=960$