Class 11th

11. Given, a₁=3

a_n=3a_{n – 1}.+2  "n>1.

Putting n=2,3,4,5 we get,

a₂=3a_{2 – 1}+2=3a₁+2=3 * 3+2=9+2=11

a₃=3a_{3 – 1}+2=3a₂+2=3 * 11+2=33+2=35

a₄=3a_{4 – 1}+2=3a₃+2=3 * 35+2=105+2=107.

a₅=3a_{5 – 1}+2=3a₄+2=3 * 107+2=321+2=323.

Hence, the first five terms of the sequence are 3,11,35,107,323.

And the series is 3+11+35+107+323+ …

10.$a_n=\frac{x(x-2)}{x+3}$ .

Put n=20,

$a_{20}=\frac{20(20-2)}{20+3}=\frac{20*18}{23}=\frac{360}{23}$

9. a_n= ( –1)^{n – 1} .n³.

Put n=9 we get,

a_q= (–1)^{9 – 1}.9³= (–1)⁸. 729=729.

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