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

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

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

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

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

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

4.32

(a) Let $\mathit{\theta }\left(\mathit{t}\right)=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{{\mathit{v}}_{\mathit{o}\mathit{y}}-\mathit{}\mathit{g}\mathit{t}}{{\mathit{v}}_{\mathit{o}\mathit{x}}}\right)$ be the angle at which the projectile is fired w.r.t. the x-axis, since ${\mathit{\theta }}_0=\mathit{}{\mathbf{tan}}^{-1}?\left(\frac{4{\mathit{h}}_{\mathit{m}}}{\mathit{R}}\right)$ depends on t

Therefore tan $\theta$ since (Vy=Voy -gt and Vx = Vox)

(b) Since $\theta$ = $\theta \left(t\right)=\frac{Vx}{Vy}=\frac{Voy-gt}{Vox}$ sin2 $\theta \left(t\right)={\tan }^{-1}?\left(\frac{Voy-gt}{Vox}\right)$ /2g…….(1)

R = $h_{max}$ sin2 $u^2$ /g …….(2)

Dividing (1) by (2)

( $\theta$ /R) = [ $u^2$ sin2 $\theta$ /2g]/[ $h_{max}$ sin2 $u^2$ /g] = $\theta$ / 4

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 

4.31

Speed of the cyclist = 27 km/h = 7.5 m/s

Radius of the road = 80m

The net acceleration is due to braking and the centripetal acceleration

Acceleration due to braking = 0.5 m/s²

Centripetal acceleration a = v²r = (7.5)2/80= 0.703 m/s²

The resultant acceleration is given by a = sqrt ( $a_t^2$ + $a_c^2$ ) = sqrt ( ${0.5}^2$ + ${0.7}^2$ ) = 0.86 m/s²

tan $\beta$ = AC / at = 0.7/0.5,  $\beta$ = 54.5 $°$

4.30

Speed of the fighter plane = 720 km/h = 200 m/s

The altitude of the plane = 1.5 km =1500 m

Velocity of the shell = 600 m/s

$\sin ?\theta$ = 200/600

$\therefore \theta =19.47°$

Let H be the minimum altitude for the plane to fly, without being hit

Using equation H = $u^2$ sin2 (90- $\theta )/2g$

= (6002cos2 $\theta$ )/2g

= 360000 { ( 1 + cos2 $\theta$ )/2}/2 $*$ 10

= 16000 m = 16 km

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

4.29 The angle of projectile $\theta$ = 30 $°$

The bullet hits the ground at a distance of 3 km, Range R = 3 km

We know horizontal range for a projectile motion, R = $u^2$ sin2 $\theta$ / g

3 = $u^2$ sin60 $°$ / g

$\frac{u^2}{g}$ = 3/ sin60 $°$ = 3.464 …………… (1)

To hit a target at 5 km,

Max Range,  $R_{max}$ = $\frac{u^2}{g}$ …………. (2)

On comparing equation (1) and (2), we get

$R_{max}$ = 3.464

Hence, the bullet will not hit the target even by adjusting its angle of projection.

4.28

(a) We cannot associate the length of a wire bent into a loop with a vector.

(b) We can associate a plane area with a vector.

(c) We cannot associate the volume of a sphere with a vector.