Sec 1 Maths Questions
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Hi all,
Please help to solve the followin Sec 1 Maths Questions:
Q1) 3,12,25,42 are the 1st four terms of a number sequence.
(a) What is the nth term of the sequence? Express in terms of n.
(b) What is the 40th term?
(c) Which term is the number 1537?
Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2
(a) Express the sum of S in terms of n in the nth line sequence.
Thanks.
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Originally posted by kkhing:
Hi all,
Please help to solve the followin Sec 1 Maths Questions:
Q1) 3,12,25,42 are the 1st four terms of a number sequence.
(a) What is the nth term of the sequence? Express in terms of n.
(b) What is the 40th term?
(c) Which term is the number 1537?
Q2) Line 1: 2 + 6 = 8 = 2 x 2^2
Line 2: 2 + 6 + 10 = 18 = 2 x 3^2
Line 3: 2 + 6 + 10 +14 = 8 = 2 x 4^2
(a) Express the sum of S in terms of n in the nth line sequence.
Thanks.3 + 9 = 12
12 + 13 = 25
25 + 17 = 42
Notice that the difference between 9 and 13, as well as 13 and 17 are both 4.
This should be enough hint for you to form your nth term. Once you formed it, the rest of that question should not be a problem.
For the second question, look at the most right side of those equations of the lines. It should give you a very big clue on how the nth line should be like in terms of n. That will be your answer.
Most of the forummers, especially the moderators, will try not to give direct answers to students. But feel free to ask again, if you are truly still stumped.
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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.
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Q1) 3,12,25,42 are the 1st four terms of a number sequence.
Divide each term by (n+2)
You will get 1,3,5,7, whose nth term is (2n-1)
So in order to get back the nth term of 3,12,25,42, just multiply (2n-1) by (n+2) which gives you (2n-1)(n+2).
For Q2, notice a pattern in the Line number and the RHS of the equation.
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Originally posted by kkhing:
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 -28.5The 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
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Originally posted by kkhing:
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.
Buy the MOE recommended Casio Fx 95 SG PLUS calculator and uses the equation mode in the calculator to find these nth term formula easily.
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Originally posted by Lee012lee:
Buy the MOE recommended Casio Fx 95 SG PLUS calculator and uses the equation mode in the calculator to find these nth term formula easily.
eeeeeee teacher, how can you teach people this kind of cheapskate method
Just joking about the above, indeed it will be much better to get an approved calculator to help out for such questions, especially during exams.
Knowing how to do the question by hand is a big plus, but in stress-packed examination conditions, cutting down on time spent per question is a priority.
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It brings back my sec school memories.