How do you find the nth term rule for #34,25,16,7,-2,...#?

1 Answer
Oct 2, 2016

Answer:

If you consider #a_0=34#, then #a_n = a_0-9n#
If you consider #a_1=34#, then #a_n = a_0-9(n-1)#

Explanation:

First of all, we need to see that the rule of the sequence is "subtract nine", and I suppose that this is the easy part.

Now, we need to see the sequence with the point of view of the first term: we start with #34#, and obtain #25# by subtracting #9#. Now, what's the link between the first and the third element? We obtain #16# by subtracting #9# again from #25#, which means that we are subtracting #9# twice from #34#.

This should lead you to the general rule: start with a number, subtract #9# once to obtain the second number, subtract #9# twice to obtain the third number...and so on.

Using a proper notation, if we define #a_0# as the first term of the sequence, we can write

#a_1 = a_0-9#
#a_2 = a_1-9 = (a_0-9)-9 = a_0-2*9#
#a_3 = a_2-9 = ( a_0-2*9)-9 = a_0-3*9#
#\vdots#
#a_n = a_0 - n*9#

which would be your closed formula for #a_n#

NB: if you choose to start with #a_1# instead of #a_0#, then you must rescale your coefficients, and you would have #a_n = a_1 - (n-1)*9#