Maths As again?...?

Question

the first three terms of an arithmetic seris are (12-p),2p,(4p-5) respectively where p is a constant.
The value of p we've found...its part a)

but how many values of the series have a value of less than 400?

a)2p-(12-p) = 4p-5-2p
3p-12 = 2p-5
p=7

Answer 1

If p = 7 then the first three terms are 5 , 14 , 23
term.......value
1............5
2...........14
3............23
nth........(9n - 4)
so if 9n - 4 < 400..... n < 44.888..
so n is 44......... there are 44 terms in this series that have a value of less than 400

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Answer 2

This is an arithmetic series, thus

2p - (12 - p) = 4p - 5 - 2p

3p - 12 = 2p - 5

p=7

Now, a(1) = 5
a(2) = 14

a(2) - a(1) = 9
a(n) = 5 + 9(n - 1) = 5 + 9n - 9 = -4 + 9n

Let's solve a(n)<400

-4 + 9n < 400

9n < 404

n < 44.8888....

The 1st 44 terms are smaller than 400

Answer 3

The terms start as 5, 14, 23, etc. The nth term is 9n-4.
So let 9n-4=400 n= just less than 45. so there are 44 values less than 400.

Answer 4

p = 5, then the series are 5, 10, 15,...

the ith term = u(i) = 5 + (i - 1) * 5

5 + (i - 1)*5 < 400
5 + 5i - 5 < 400
5i < 400
i < 80

when i = 80, the ith term will be 400. then the number of values that is less than 400 is 80 - 1 = 79.