Answer: Option A. -> 5th term
:
A
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 APis8and 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 to0.

Answer: Option C. -> 24
:
C
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
$t_n=a+(n-1)d.$
$\Rightarrow t_5=4+4\left(5\right)=24$

Answer: Option D. -> 672
:
D
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 is5, 7, 9, 11 . . .
The sum to $n$ terms of an AP with first term $a$ and common difference $d$ is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right].$
$\therefore S_{24}=\frac{24}{2}\left(2\right(5)+(24-1\left)2\right)$
$=12(10+46)=672$

Answer: Option B. -> 960
:
B
The first 15 multiples of 8 are8, 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 findsum of $n$ terms of AP,we get
$S_{15}=\frac{15}{2}\left(2\right(8)+(15-1\left)\right(8\left)\right)$
$=\frac{15}{2}(16+8(14\left)\right)$
$=\frac{15}{2}(16+112)$
$=\frac{15}{2}\left(128\right)=960$

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

Answer: Option A. -> 34, 27
:
A
(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.

:
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 = $\pi \left(45.5\right)=143cm$

Answer: Option B. -> a+b2
:
B
The arithmetic mean of two numbers is the simple average of the two numbers.
Thus,
$AM=\frac{\text{Sum of the numbers}}{\text{Number of numbers}}$
$=\frac{a+b}{2}$

Answer: Option A. -> n2(13−3n)
:
A
The given sequence is5, 2, -1, -4, -7, ...., an APwith 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
$S_n=\frac{n}{2}[2a+(n-1\left)d\right].\Rightarrow S_n=\frac{n}{2}[10+(n-1\left)\right(-3\left)\right]$
$=\frac{n}{2}[10-3n+3]$
$=\frac{n}{2}[13-3n]$

Answer: Option D. -> 13
:
D
Here
First term, a = 7
Common difference, d = 10 - 7 = 3
Using the formula for $n^{th}$ term
$a_n=a+(n-1)d$
$43=7+(n-1)3$
$36=(n-1)3$
$12=n-1$
$n=13$
There are total 13 terms in the given AP.