Class 11th

62. Since the numbers of bacteria doubles every hour. The number after every hour will be a G.P

So, a=30

r=2

$\therefore$ At end of 2^nd hour, a₃ (or 3^rd term) = $a{r}^2=30*2^2$

= 30*2⁴

= 120

At end of 4^th hour, a₅ (r 4^th  term) = $a{r}^4$

= 30*2⁴

= 30*16

= 480

Following the trend,

And at the end of nth hour, an+1= $a{r}^{n+1-1}=a{r}^n$

= 30 *2ⁿ

4.27 No in the both the cases.

A physical quantity which is having both direction and magnitude is not necessarily a vector. For instance, in spite of having direction and magnitude, current is a scalar quantity. The basic necessity for a physical quantity to fall in a vector category is that it ought to follow the “law of vector addition”.

As the rotation of a body about an axis does not follow the basic necessity to be a vector i.e. it does not follow the “law of vector addition”.

4.26 Yes and No

A vector in space has no distinct location. The reason behind this is that a vector stays unchanged when it displaces in a way that its direction and magnitude do not change.

A vector changes with time. For instance, the velocity vector of a ball moving with a specific speed fluctuates with time.

Two equivalent vectors situated at different locations in space do not generate the same physical effect. For instance, two equivalent forces acting at different points on a body to rotate the body, but the combination will not generate the equivalent turning effect.

61. Let a and b be the two number and a> b  so a-b = (+ ve)

60. Let a and b be the two numbers and a>b so a-b = (+ ve)

So, sum of two numbers = 6. G.M of a and b

59. We know that,

G.M between a and b = √ab

Given $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}=\; \surd ab$

$\Rightarrow$ $a^{n+1}+b^{n+1}={\left(ab\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(a^n+b^n\right)$

$\Rightarrow$ $a^{n+1}+b^{n+1}=a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{b}^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\Rightarrow$ $a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]=b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$\Rightarrow$ $\frac{a^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{b^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}{\left[a^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-b^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}$

$\Rightarrow$ ${\left(\frac{a}{b}\right)}^{n+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=1.={\left(\frac{a}{b}\right)}^{?}$

$\Rightarrow$ $n+\frac{1}{2}=0$

$\Rightarrow$ $n=-\frac{1}{2}$

58. Let G₁ and G₂ be the two numbers between 3 and 81 so that 3, G₁, G₂, 81 is in G.P.

So, a = 3

a₄ = ar³ = 81 (when r = common ratio)

$\Rightarrow$ $r^3=\frac{81}{a}=\frac{81}{3}$

$\Rightarrow$ r³ = 27

$\Rightarrow$ r³ = 3³

$\Rightarrow$ r = 3

So, G1 = ar = 33=9 and G₂ = ar² 3´ (3)² =27

57. Let r be the common ratio of the G.P. then,

First term = a₁ = a = a

$a_2=ar=b$

$a_3=a{r}^2=c$

$a_4=a{r}^3=d$

So, L.H.S.= $\left(a^2+b^2+c^2\right)\left(b^2+c^2+d^2\right)$

= $\left[a^2+{\left(ar\right)}^2+{\left(a{r}^2\right)}^2\right]\left[{\left(ar\right)}^2+{\left(a{r}^2\right)}^2+{\left(a{r}^3\right)}^2\right]$

= $\left[a^2+a^2{r}^2+a^2{r}^4\right]\left[a^2{r}^2+a^2{r}^4+a^2{r}^6\right]$

= $a^2\left[1+r^2+r^4\right]a^2{r}^2\left[1+r^2+r^4\right]$

= $a^4{r}^2{\left[1+r^2+r^4\right]}^2$

R.H.S.= ${\left(ab+bc+cd\right)}^2$

= ${\left[a*ar+ar.a{r}^2+a{r}^{2}a{r}^3\right]}^2$

= ${\left[a^{2}r+a^2{r}^3+a^2{r}^5\right]}^2$ $$

= { $a^{2}r{\left[1+r^2+r^4\right]}^2$ }

= $a^4{r}^2{\left[1+r^2+r^4\right]}^2$

L.H.S. = R.H.S.

56. Let a and r be the first term and common difference of the G.P

Thus, Sum of the first on term, $S_n=\frac{a\left(1-r^n\right)}{1-r}$

Let Sn = sum of term from (n+1)^th to (2n)^th  term

= $a{r}^n+a{r}^{n-1}+......+a{r}^{2n-1}$

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

[ $?$ the above is a G.P. with first term ar^x and common ratio = $\frac{a{r}^{n+1}}{a{r}^n}=r^{n+1-n}=r$

and number of term from (2n)^th to (n+1)^th = n

So, $\frac{S_n}{S_n}=\frac{\frac{a\left(1-r^n\right)}{1-r}}{\frac{a{r}^n\left(1-r^n\right)}{1-r}}$

= $\frac{a}{a{r}^n}$

= $\frac{1}{r^n}$

55. Given, first term and xth term and a and b

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

Then,

Product of x terms, p= $\left(a\right)$ $\left(ar\right)$ $\left(a{r}^2\right)$ …… $\left(a{r}^{n-1}\right)$

p= $\left(a\right)$ $\left(ar\right)\left(a{r}^2\right)$ …. $\left(a{r}^{n-2}\right)\left(a{r}^{n-1}\right)$

= (a *a* ….n term)(r´r²´r³ …. $r^{n-2}*r^{n-1}$ )

= $a^x$ $r^{\left[1+2+3+....\left(n-2\right)+\left(n-1\right)\right]}$

p = $a^n$ $r^{n\left(\frac{n-1}{2}\right)}$

So, $p^2={\left[a^n{r}^n\left(\frac{n-1}{2}\right)\right]}^2$ [We know that,

= $n\frac{\left(n-1\right)}{2}$ [ $1+2+3+...+n=\frac{\left(n-1\right)n}{2}$

= $a^{2n}{r}^{2n}\left(\frac{n-1}{2}\right)$ [So, $1+2+3+....+\left(n-1\right)=\frac{\left(n-1+1\right)\left(n-1\right)}{2}=\frac{n(n-1)}{2}$

= $a^{2n}*r^{n\left(n-1\right)}$

= ${\left(a*a{r}^{n-1}\right)}^n$ $[?$ $a{n}^{n-1}=$ last term = b (given)]

= ${\left(a*b\right)}^n$