Question

write in set builder form where B= {0,3,8,15......99}
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Answer 1

Answer:

We have,

B= {0, 3, 8, 15, ......, 99}

If you carefully look towards the elements of the set B, you'll find out that, all these elements are just one less than the squares of natural numbers from 1 to 10. i.e.,

1^2 - 1 = 0

2^2 - 1 = 4 - 1 = 3

3^2 - 1 = 9 - 1 = 8

4^2 - 1 = 16 - 1 = 15

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10^2 - 1 = 100 - 1 = 99.

So, in set builder form, B can be shown as

B = { x^2 : x€N & x<11}

Step-by-step explanation:

Answer 2

Answer:

Let's take the difference between every term and the term succeeding it

i.e.,

3-0=3,

8-3=5,

15-8=7,

24-15=9,

35-24=11,

48-35=13,

63-48=15

80-63=17

99-80=19......

These differences form a sequence which is in Arithmetic Progression (A.P) with the first term being 3, and the common difference being 2.

The sequence is 3,5,7,9,11,...

So obviously the next term would be 13 i.e., the difference between say 'x' and 35 is 13.

a-35=13

a=35+13=48

Similarly,

b-48=15

b=48+15=63

Similarly,

c-63=17

c= 17+63= 80

Similarly,

d-80=19

d= 19+80=99

:)

So the original sequence would be 0,3,8,15,24,35,48,63,...

or

1X1-1: 0

2X2-1: 3

3X3-1: 8

4X4-1: 15

5X5-1: 24

6X6-1: 35

7X7-1: 48

8X8-1: 63

9x9-1: 80

10x10-1: 99

Step-by-step explanation:

i hope it helps you