Class 11th

8. $a_n=\frac{n^2}{2^n}$

Put n=7,

$a_7=\frac{7^2}{2^7}=\frac{49}{128}$

7. a_n = 4_n – 3.

Putting n = 17, we get

a₁₇ = 4 * 17 – 3=68 – 3 = 65.

and putting n = 24 we get,

a₂₄ = 4 * 24 – 3 = 96 – 3 = 93

5. Here, a_n= (–1)^{n – 1} . 5ⁿ⁺¹

Putting n=1,2,3,4,5 we get,

a₁= (–1)^{1 – 1}.5¹⁺¹= (–1)⁰* 5²=25

a₂= (–1)^{2 – 1}.5²⁺¹= (–1)¹* 5³= –125

a₃= (–1)^{3 – 1}. 5³⁺¹= (–1)²* 5⁴=625

a₄= (–1)^{4 – 1}. 5⁴⁺¹= (–1)³. 5⁵= –3125

a₅= (–1)^{5 – 1}. 5⁵⁺¹= (–1)⁴. 5⁶=15625.

Hence, the first five terms are 25, –125,625, –3125 and 15625.

4. Here, $a_n=\frac{2n-3}{6}$

Putting n=1,2,3,4,5 we get,

$a_1=\frac{2*1-3}{6}=\frac{2-3}{6}=-\frac{1}{6}$

$a_2=\frac{2*2-3}{6}=\frac{4-3}{6}=\frac{1}{6}$

$a_3=\frac{2*3-3}{6}=\frac{6-3}{6}=\frac{3}{6}=\frac{1}{2}$

$a_4=\frac{2*4-3}{6}=\frac{8-3}{6}=\frac{5}{6}$

$a_5=\frac{2*5-3}{6}=\frac{7}{6}$

Hence, the first five terms are $-\frac{1}{6},\frac{1}{6},\frac{1}{2},\frac{5}{6},and\frac{7}{6}$

1. Here a_n = n (n + 2)

Substituting n=1,2,3,4,5 we get,

a₁=1 (1+2)=1 * 3=3

a₂=2 (2+2)=2 * 4=8

a₃=3 (3+2)=3 * 5=15

a₄=4 (4+2)=4 * 6=24

a₅=5 (5+2)=5 * 7=35.

Hence, the first five terms are 3,8,15,24 and 35

27. Since the parabola is symmetric with respect to y-axis and has vertex (0, 0)

The equation in of the form x³ = 4ay or x² = 4ay.

The parabola passes through (5, 2) which lies on the 1st quadrant

? The equation of parabola is of the form,

x² = 4ay

Putting x = 5 and y = 2,

(5)² = 4 (a) (2)

25 = 8a

a = $\frac{25}{8}$

? The equation of the parabola is,

x² = 4ay

$\Rightarrow x^2=4\left(\frac{25}{8}\right)y$

$\Rightarrow x^2=\frac{25}{2}y$

2x² = 25y

26. Since the axis of parabola is x-axis,

The equation parabola is either y² = 4ax or y² = 4ax.

Also it passes through (2, 3) which lies in the first quadrant.

So the equation is,

y²= 4ax

Putting x = 2 and y = 3, we set

(3)² = 4 (a) (2)

a = $\frac{9}{8}$

? The equation of parabola is y² = 4 $*\frac{9}{8}x$

y² = $\frac{9}{2}x$

2y² = 9x

25. Focus (-2, 0) lies on x-axis and the x-Co-ordinate is negative.

The equation must be, y² = 4ax

Co-ordinate of focus = (–a, 0) = (–2, 0)

–a = –2

a = 2

? Equation of a parabola is,

y² = –4 (2)x

y² = –8x

24. Since the focus (3, 0) lies on the x-axis, the x-axis is the axis of parabola.

Hence the equation is either y² = 4ax or y² = –4ax

Since the focus has positive x co-ordinate,

The equation must be y² = 4ax

Co-ordinate of focus (a, 0) = (3, 0)

a = 3

? The equation is given by

y² = 4 (3)x

y² = 12x

23. Since the focus (0, –3) lies on the y–axis, the y–axis is the axis of parabola.

Hence the equation is either x² = 4ay or x² = –4ay

Since the directrix is y = 3 and the forces (0, –3) has negative y Co–ordinate.

The equation must be

x² = –4ay

Co–ordinate of focus = (0, –a)

(0, –a) = (0, –3)

–a = –3

a = 3

? Equation of parabola is

x² = –4ay

x² = –4 (3)y

x² = –12y