Maths 1 Marks Question

Let $\mathrm{A}$ be the event that the number obtained on the upper face is even.

$\therefore A=\{2,4,6\}$

$S=\{1,2,3,4,5,6\}$

$\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}$

$\therefore \text { Common difference }=t_2-t_1$

= 0.6 - 0.9

= -0.3

$\therefore t_1=3(1)-2$

$=3-2$

$=1$

$t_2=3(2)-2$

$=6-2$

$=4$

$\therefore$ The first two terms of the sequence are 1 and 4.

$\therefore t_1=2(1)+1$

= 2 + 1

= -3

$\therefore$ The first term of the sequence is 3.

$\therefore t_1=2(1)-5$

= 2 - 5

= -3

$t_2=2(2)-5$

= 4 - 5

= -1

$\therefore$ The first two terms of the sequence are -3 and -1 .

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

where $a=$ first term,

$d=$ common difference

OR

$S_n=\frac{n}{2}\left(t_1+t_n\right)$

where $t_1=$ first term,

$t_n=\text { last term }$

$\therefore$ First term $(a)=t_1=1$ and

Common difference $(d)=t_2-t_1$

$=4-1$

$=3$

$\therefore \mathrm{t}_2-\mathrm{t}_1=4-2=2$

$\mathrm{t}_3-\mathrm{t}_2=6-4=2$

$\therefore \mathrm{t}_2-\mathrm{t}_1=\mathrm{t}_3-\mathrm{t}_2=\ldots=2=\text{constant}$

The difference between two consecutive terms is constant.

$\therefore$ The given sequence is an A.P.

If $b^2-4 a c<0$, then the roots are not real.

$\therefore p-5=0 \text { or } p+3=0$

$\therefore p=5 \text { or } p=-3$

$\therefore$ The roots of the given equation are 5 and -3 .

$\therefore 4 y^2-3 y+7=0$

Comparing the above equation with $a y^2+b y+c=0$, we get $a=4, b=-3, c=7$

When $b=4, a=-3+4=1$

$\therefore(-2,1)$ and $(1,4)$ are the solutions of $a-b=-3$.

$=5(0)+3(2)$

$=0+6$

$=6$

$=\text { R.H.S }$

$\therefore(0,2)$ is the solution of the given equation.

When $x=3, y=7-3=4$

$\therefore(1,6)$ and $(3,4)$ are the solutions of $x+y=7$.

4(0)+5 y=20

$\therefore 5y=20$

$\therefore y=\frac{20}{5}$

= 4

OR

$y-x=3 \text { where } y>x$