Ncert Solutions Maths class 11th

67. The given series is 3 * 1² + 5 * 2² + 7 * 3² + ….

So,a_n = (n^th term of A P 3, 5, 7, .) (n^th term of A P 1, 2, 3, ….)²

a = 3, d = 5 -3 = 2a = 1, d = 2 -1 = 1.

= [3 + (n- 1) 2] [1 + (n- 1) 1]²

=[3 + 2n- 2] [1 + n- 1]²

(2n + 1)(n)²

= 2n³ + n²

So, = 5n2∑n³ + ∑n²

$=2·{\left[\frac{n^{}(n+1)}{2}\right]}^2+\left[\frac{n(n+1)(2n+1)}{6}\right]$

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

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

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

$=\frac{n}{2}(n+1)·\left[\frac{3n^2+5n+1}{3}\right]$

$\frac{n(n+1)\left(3n^2+5n+1\right)}{6}$

66. Given series is 1* 2* 3 + 2* 3 *4 + 3* 4 *5 + … to n term

a_n = (n^th term of A. P. 1, 2, 3, …) ´* (n^th terms of A. P. 2, 3, 4) *

i e, a = 1, d = 2- 1 = 1i e, a = 2, d = 3- 2 = 1

(nth term of A. P. 3, 4, 5)

i e, a = 3, d = 3 -4 = 1.

= [1 + (n -1) 1] *[2 + (n -1):1]* [3 + (n- 1) 1]

= (1 + n -1)*(2 + n -1)*(3 + n -1)

= n (n + 1)(n + 2)

= n(n² + 2n + n + 2)

=n³ + 2n² + 2n.

S_n = ∑n³ + 3 ∑n² + 2 ∑n

$={\left[\frac{m(n+1)}{2}\right]}^2+\frac{3n(n+1)(2n+1)}{6}+\frac{2n(n+1)}{2}$

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

= $\frac{n(n+1)}{2}\left[\frac{n^2+n}{2}+2n+1+2\right]$

$=\frac{n(n+1)}{2}\left[\frac{n^2+n+2*2n+2*3}{2}\right]$

$=\frac{n(n+1)}{2}\left[\frac{x^2+n+4x+6}{2}\right]$

$=\frac{n(n+1)}{2}*\frac{\left[n^2+5n+6\right]}{2}$

$=\frac{n(n+1)}{4}\left[n^2+2n+3n+6\right]$

$=\frac{n(n+1)}{4}[n(n+2)+3(n+2)]$

$=\frac{n(n+1)(n+2)(n+3)}{4}$

65. Given series is 1*2+2 *3+3* 4+4* 5+…

So, an (nth term of A.P 1, 2, 3…) (nth term of A.P. 2, 3, 4, 5…)

i e, a = 2, d = 2 -1 = 1i e, a = 2, d = 3 - 2 = 1

= [1 + (n- 1) 1] [2 + (n -1) 1]

= [1 + n- 1] [2 + n -1]

= n (n -1)

= n²-n.

S_n (sum of n terms of the series) = ∑n² + ∑n.

S_n = $\frac{n(n+1)(2n+1)}{6}$ + $\frac{n(x+1)}{2}$

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

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

$=\frac{n(n+1)}{2}*\frac{(2n+4)}{3}$

$=\frac{n(n+1)*2(n+2)}{2*3}$

$=\frac{n(n+1)(n+2)}{3}$

64. Let a and b be the roots of quadratic equation

So, A.M = 8

$\Rightarrow$ $\frac{a+b}{2}=8$ a + b =16 ….I

G.M. = 

$\Rightarrow$ ab = 25…. II

We know that is a quadratic equation

$x^2-x$  (sum of roots) + product of roots = 0

$x^2-x\left(a+b\right)+ab=0$

$x^2-16x+25=0$ using I and II

Which is the reqd. quadratic equation 

63. Given,

Principal value, amount deposited, P= ?500

Interest Rate, R= 10

Using compound interest = simple interest + $\frac{P*R*time}{100}$

Amount at the end of 1st year

= $500+\frac{500*10+1}{100}=500\left(1+0.1\right)=500*1.1$

= ?500* (1.1)

Amount at the end of 2nd year

= $500\left(1.1\right)+\frac{500*1.1*10*1}{100}$

= $500\left(1.1\right)\left(1+0.1\right)$

= ?500 (1.1)²

Similarly,

Amount at the end of 3rd year = ?500 (1.1)³

So, the amount will form a G.P.

? 500 (1.1)? 500 (1.1)²?500 (1.1)³, ……….

$\therefore$ After 10 years = ?500 (1.1)¹⁰

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

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