# Generate the first 5 terms of each sequence: BRONZE In the form n2 + b 1) n2 + 1 2) n2 + 3 3) n2 + 10 4) n2 - 1 5) n2 - 14 SILVER In the form an2 + b 1) 2n2 + 5 2) 2n2 + 7 3) 2n2 - 4 4) 3n2 +...

Generate the first 5 terms of each sequence:

BRONZE In the form n2 + b

1) n2 + 1

2) n2 + 3

3) n2 + 10

4) n2 - 1

5) n2 - 14

SILVER In the form an2 + b

1) 2n2 + 5

2) 2n2 + 7

3) 2n2 - 4

4) 3n2 + 5

5) 3n2 - 4

Part B Nth TERM OF QUADRATIC SEQUENCES

Find the nth term of the quadratic sequences:

BRONZE In the form n2 + b

1) 3, 6, 11, 18, 27

2) 7, 10, 15, 22, 31

3) 8, 11, 16, 23, 32

4) 0, 3, 8, 15, 24

5) -8, -5, 0, 7, 16

SILVER In the form an2 + b

1) 11, 17, 27, 41, 59

2) 7, 16, 31, 52, 79

3) 6, 18, 38, 66, 102

4) 0, 9, 24, 45, 72

5) 0, 12, 32, 60, 96

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### 1 Answer

It is not clear from the question if it is n2 or n^2. I am assuming it is n^2.

Bronze series: first 5 terms

- n2 + 1: take n = 1, we get 2. Similarly for n= 2,3,4,5; we get 5,10,17,26.
- Similarly, for n2+3, we can substitute n as 1,2,3,4,5 and obtain the series as 4,7,12,19,28.

others in the bronze series can be solved similarly.

Silver series:

- 2n2 + 5: we can substitute n as 1,2,3,4,5; and obtain 7,13,23,37,55.
- 2n2+7: we can similarly substitute n= 1,2,3,4,5 and obtain the series as 9,15,25,39,57.

The rest of the series can be similarly solved.

Finding the n the term in series:

3,6,11,18,27: each number is separated by 2 and if we look carefully this is a series of n2+2.

Similarly, next series is n2+6.

Other parts can also be solved similarly.

Hope this helps.