The problem is I'm not the student but the parent of a sec 1
student. Can you show me how you solve the problem. This type of
question was not taught during my years of study when I was in the
sec school time. Thanks.
Sorry about the hint I have given, I just realise it is not
enough for a Sec 1 student, much less to say his/her parent. To be
honest, the first question can be very challenging for lower sec
students, especially when there are no hints.
I am going to show you how to do it, but I think I will need
some forummers to help breakdown some of my explanations for easier
understanding.
Notice that we are adding 9, 13, 17 (the next few will be 21,
25, 29...) and the difference between each of them is constantly
4.
Thus, the nth term will have to be a second order expression
(Something of the form an^2 + bn + c, where a,b and c are all
integers).
Now we will need to do some substitution:
3 = a(1)^2 + b(1) + c = a + b + c -----(1)
12 = a(2)^2 + b(2) + c = 4a + 2b + c -----(2)
25 = a(3)^2 + b(3) + c = 9a + 3b + c -----(3)
From (1) and (2), we will have 3a + b = 9 -----(4)
From (2) and (3), we will have 5a + b = 13 ------(5)
From (4) and (5), we will have a = 2
Since a = 2, using (4), we know that b = 3.
Using (1), and the fact that a = 2, b = 3, we know that c =
-2
Substituting everything back in, we know the nth term is
2n^2 + 3n -
2
For (b), the 40th term will be 2(40)^2 + 3(40) - 2 =
3318
For (c), let 1537 = 2n^2 + 3n - 2.
2n^2 + 3n - 1539 = 0
n = [-3 +/- sqrt (3^2 -4(2)(-1539)] / 2(2)
= [-3 +/- sqrt(12321)] / 4
= [-3 +/- 111] / 4
= 108/4 o r -114/4
= 27 or -28.5
The term cannot possibly be negative, thus the answer must be
the 27th term.
For the second question, notice that:
Line 1: 2 x 2^2
Line 2: 2 x 3^2
Line 3: 2 x 4^2
Thus the nth line will be 2 x (n + 1)^2 = 2(n + 1)^2
