In figure n, length is n, width is 2n + 1, so area is n (2n + 1)

So for figure 150, n is 150, 2n + 1 is 301 so area is 150 x 301 = 45150 cm2.

So what do I do with the unknown value ?

We would need to substitute n = 150 into the ‘n’th term to find the value. It’s easier to find a pattern for the height and the width rather than manually finding it by counting.

Ohhh ... okay ,Thank you :)

But why 150 times 2?

If the number is squares is 6105, then their total area is 6105 cm2.

n (2n + 1) = 6105

Either we do trial and error or we use quadratic factorisation.

Trial and error: if n = 50, 2n + 1 is 101, 50 x 101 = 5050 so not yet

Try n = 55, 2n = 111 and 55 x 111 is 6105 so yes, n = 55 so this is figure 55.

Quadratic factorisation

n (2n + 1) = 6105
2n2 + n - 6105 = 0
(yes, 6105 is a very large number)
(2n + 111) (n - 55) = 0
n cannot be negative so n = 55

Width is 2n + 1

2 times 4 plus 1 = 9 for figure 4

How to see it’s 2n + 1?

3, 5, 7, 9, ...

The difference in term is 2. We put this in front of n as ‘2n’.

The ‘zero’th term is 1 since the first term is 3. We put this as 1.

Finally, we add the two together to get 2n + 1.

Oh yes . Thank you so much for your help :)

It’s actually 150 x 2 + 1. Notice that the width is almost twice the height, but a bit more than that.

No problem!

Solving quadratic equations is not taught yet in P6 so guess and check is the way to go for c)

I don’t recall primary school dealing with general nth term for these type of questions though; this question looks more like a secondary question.

Yes but that was last time. Now they expect P6 to learn how to solve patterns found in Sec1 E Math, but using the idea of figure/pattern number instead of algebraic terms like n, x or a

ie to deduce from the question that
height = figure number,
width = 2 x figure number + 1