How do you write the first five terms of the sequence defined recursively #a_1=25, a_(k+1)=a_k-5#, then how do you write the nth term of the sequence as a function of n?

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Answer 1 · 2017-07-17T21:21:33

We obtain general form for the equation of Arithmetic Progression from the reference:

#a_n = a_1 + (n-1)d#

We are given #a_1 = 25# and the equation #a_(k+1) = a_k-5# implies that #d = -5#, therefore, the equation for the nth term is:

#a_n = 25 + (n-1)(-5); {n in ZZ|n>0}#

Answer 2 · 2017-07-17T23:02:46

#a_k = 30 - 5k#, #k = 1, 2, . . . , n#

#=> ul(a_n = 30 - 5n)#

Well, the #a_1# term is given as #25#, and the successive terms #a_(k+1)# are five less than the current term, #a_k#. So, the sequence given is:

#25, 20, 15, 10, 5, 0, -5, . . . #

This is very similar to the graph #y = -5x#; the term index is represented by the #x# value corresponding to the term represented by the #y# value:

graph{-5x [-30.45, 42.63, -7.54, 28.97]}

In the graph, we analogously have:

#y = 25, 20, 15, 10, 5, 0, -5, . . . #
#x = -5, -4, -3, -2, -1, 0, 1, . . . #

This means the #k#th term, iterating #k# from #1# to #n#, can be found using the series:

#color(blue)(a_k = 30 - 5k, k = 1, 2, 3, . . . , n)#