Class 11th

41. Here,  $a_1=1$

$r=\frac{-9}{1}=-a$

So,  $s_n=\frac{a_1\left[1-r^n\right]}{1-r}$

= $\frac{1\left[1-{\left(-a\right)}^n\right]}{1-\left(-a\right)}$

= ${\frac{1-\left(a\right)}{1+a}}^n$

39. Here a = 0.15 = $\frac{15}{100}$

$r=\frac{0.015}{0.15}=\frac{\frac{15}{1000}}{\frac{15}{150}}=\frac{1}{10}$ <1

n = 20

So, Sum to term of G.P., $s_n=\frac{a(1-r^n)}{1-r}$

$\Rightarrow$ $s_{20}=\frac{\frac{15}{100}\left[1-{\left(\frac{1}{10}\right)}^{20}\right]}{1-\frac{1}{10}}$

= $\frac{\frac{15}{100}\left[1-{(0.1)}^{20}\right]}{\frac{9}{10}}$

= $\frac{15}{100}*\frac{10}{9}\left[1-{(0.1)}^{20}\right]$

= $\frac{1}{6}\left[1-{(0.1)}^{20}\right]$

62. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n (H) = 100

Number of students who know English = n (E) = 50

Number of students who know both English & Hindi = 25 = n (H$\cap$E)

As each of students knows either Hindi or English,

Total number of students in the group,

n (H$\cup$E) = n (H) + n (E) - n (H$\cap$E)

= 100 + 25

= 125,

61. Let T and C be sets of students taking tea and coffee.

Then, n (T) = 150, number of students taking tea

n (C) = 225, number of students taking coffee

n (T$\cap$C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

n (T$\cup$C) = n (T) + n (C) n (T$\cap$C)

= 150 + 225 100

= 275

Number of students taking neither tea coffee

= Total number of students No of students taking either tea or coffee

= 600 275

= 325.

60. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A$\cap$B = {x, y}$\cap$ {y, z} = {y}≠$?$

B$\cap$C = {y, z} $\cap$ {x, z} = {z}≠$?$

A$\cap$C = {x, y}$\cap$ {x, z} = {x}≠$?$

But A$\cap$B$\cap$C = (A$\cap$B)$\cap$ C

= {y} $\cap$ (x, z}

=$?$

59. Let A, B and x be sets such that,

A$\cap$x = B$\cap$x =$?$ and A$\cup$x = B$\cup$x.

We know that,

A = A$\cap$ (A$\cup$x)

= A$\cap$ (B$\cup$x) [? A$\cup$x = $\cup$Bx]

= (A$\cap$B) (A$\cap$x) [by distributive law]

= (A$\cap$B) ∪? [? A∩x =? ]

=> A = A∩ B [? A ∪? = A]

And B = $\cap$ (B$\cup$x)

= B$\cap$ (A$\cup$x) [? B$\cup$x = A$\cup$x]

= (B$\cap$A)$\cup$ (B$\cap$x) [By distributive law]

= (B$\cap$A) ∪$?$ [? B$\cap$x =? ]

B = B$\cap$A [? A$\cup$ $?$= A]

So, A = B = A$\cap$B.

58. Let A = {a}, B = {a, b}, C = {a, c}

So, A$\cap$B = {a} $\cap$ {a, b} = {a}

A$\cap$C = {a} $\cap$ {a, c} = {a}

i.e., A$\cap$B = A$\cap$C = {a}

But B ≠C. as b$\cancel{\in }$B but $b\cancel{\ne }C$ vice-versa