Class 11th

12. When a coin is thrown we have the possible outcome of a head or a tail. And when a dice is thrown we have the possible outcomes 1, 2, 3, 4, 5 and 6. So, the desired sample space is

S = {T, (H, 1), (H, 2, 1), (H, 2, 2), (H, 2, 3), (H, 2, 4), (H, 2, 5), (H, 2, 6), (H, 3), (H, 4, 1), (H, 4, 2), (H, 4, 3), (H, 4, 4), (H, 4, 5), (H, 4, 6), (H, 5), (H, 6, 1), (H, 6, 2), (H, 6, 3), (H, 6, 4), (H, 6, 5), (H, 6, 6)}

38. For $\frac{-2}{7}$ ,  $x$ ,  $\frac{-7}{2}$ to be in G.P. we have the condition,

$\Rightarrow$ $\frac{x}{-2/7}=\frac{-7/2}{x}$

$\Rightarrow$ $x^2=\left(\frac{-7}{2}\right)*\left(\frac{-2}{7}\right)$

$\Rightarrow$ $x^2=1$

$\Rightarrow$ $x=+1$

36. Given, a = 3

Let r be the common ratio of the G.P.

Then, a₄ = (a₂)²

$\Rightarrow$ar^4-1 =   (ar^2-1)²

$\Rightarrow$ ar³=  (ar)²

$\Rightarrow$ ar³=  a²r²

$\Rightarrow$ $\frac{r^3}{r^2}=\frac{a^2}{2}$

$\Rightarrow$ r = a = 3

$\therefore$ a₇ = ar^7-1  = ar⁶= (-3) (-3)⁶  = (-3)⁷ = -2187.

35. Let a and r be the first term and the common ratio between G.P.

So, a₅= p $\Rightarrow$ a r^t-1 $$ $$ = p $\Rightarrow$ ar⁴ = p

a₈ = q $\Rightarrow$ a r^8-1 =q $\Rightarrow$ ar⁷= q

a₁₁= r $\Rightarrow$ ar^11-1 = r $\Rightarrow$ ar¹⁰= 5

So, L.H.S. q^{2 =}  (ar⁷)² = a²r¹⁴

R.H. S. = p.s= (ar⁴) (ar)¹⁰ = a¹⁺¹ r⁴⁺¹⁰  = a²r¹⁴

$\therefore$ L.H.S. = R.H.S.

34. Given, r=2

Let a be the first term,

Then, a₈= 92

$\Rightarrow$ ar^8-1=192

$\Rightarrow$a (2)⁷ = 192

$\Rightarrow$ a = $\frac{192}{2^7}=\frac{192}{128}=\frac{3}{2}$

so, a = $\frac{3}{\begin{array}{l}2\\ \end{array}}$

$\therefore$ a₁₂ = ar^12-1= $\begin{array}{l}\left(\frac{3}{2}\right)\\ \end{array}$ (2)¹¹ = 3 2^11-1

= 3*2¹⁰

= 31024

= 3072.

33. Here, a= $\frac{5}{2}$

=  $r=\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{2}$

so, a_n =  ar^n-1

i.e  a₂₀=ar^20-1= $\left(\frac{5}{2}\right)*{\left(\frac{1}{2}\right)}^{19}=\frac{5}{2}*\frac{1}{2^{19}}=\frac{5}{2^{19+1}}=\frac{5}{2^{20}}$

and an= ar^n-1= $\left(\frac{5}{2}\right)*{\left(\frac{1}{2}\right)}^{n-1}=\frac{5}{2}*\frac{1}{2^{n-1}}=\frac{5}{2^{n-1+1}}=\frac{5}{2^n}$

31. Since the man starts paying installmentas? 100 and increases? 5 every month.

a=100

d=5.

So, the A.P is 100,105,110,115, ….

? Amount of 30thinstallment=30th term of A.P =a₃₀

=a+ (30 – 1)d

=100+29 * 5.

=100+145

= ?245

i.e., He will pay? 245 in his 30th instalment.