Good thing: lots of practise of following an algorithm, and
of multiplying by 5, 7 or 9 mentally.
Bad thing: while some questions have 5 numbers (‘addends’!),
some 7 and others 9, all of the series go up by 2 each time (the method
actually works for any linear sequence).
This is a missed opportunity if only used to answer these
questions in the way stated.
Why does this method work?
This is a nice example of where algebra isn’t enough by itself: this is
so much easier to see and to understand if you choose you algebra in a particular
way. Make the middle number in
the series be n, and this is now
straightforward. (To be clear, it still works if you make the first number n, but this is more difficult to generalise from.)
With a 5-number series that increases by 2 we have: (n-4)
+ (n-2) + n + (n+2) + (n+4)
When we sum this we get 5n,
which is 5 times the middle number (as per the algorithm). But we can see why this works: there is some
symmetry in that the number to the left of n
is 2 below n while the number to the
right of n is 2 above n, so these cancel each other out. If you have additional numbers then they will
also cancel each other out, leaving only extra ns.
Can we do this if the series increases by 3? Yes – that gives us: (n-6) + (n-3) + n + (n+3)
+ (n+6)
What if the series increases by a? (n-3a) + (n-2a)
+ (n-a) + n + (n+a)
+ (n+2a) + (n+3a)
What if there is an even number of terms?
This still works (although the 'middle' is no longer an integer.
What if there is an even number of terms?
This still works (although the 'middle' is no longer an integer.
We can also use a visual method for this:
The first diagram shows 6+8+10+12+14. In the second diagram that has been converted
to 5 lots of the middle number.
This diagram also shows neatly that we are working out the
mean of the numbers in the sequence (the middle number is the mean) and then
multiplying by the number of terms.
As a final different method, it would be nice to show this
on a 100-square to illuminate further what is going on.
1 comment:
Mike Pitt (@mhpitt) makes an additional suggestion:
Can we find sequence that sums to (value)?
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