If an arithmetic sequence is added to a geometric sequence, term by term, is the resulting sequence arithmetic or geometric?

1 Answer
Aug 10, 2015

Answer:

Typically neither. The exceptions are when the arithmetic sequence is constant #0# and/or the geometric sequence has common ratio #1# (i.e. it is a constant sequence).

Explanation:

Let the arithmetic sequence be defined by #a_n = a + kn# for #n = 0, 1, 2,...#

Let the geometric sequence be defined by #b_n = b*r^n# for #n = 0, 1, 2,...#

Then the result of adding these sequences has general term:

#c_n = a + kn + b*r^n#

Arithmetic sequence ?

#c_2 - c_1 = (a+2k+br^2)-(a+k+br) = k + b(r^2-r)#

#= k + br(r-1)#

#c_1 - c_0 = (a+k+br)-(a+b) = k + b(r - 1)#

If this is an arithmetic sequence, then these two expressions must be equal, so:

#color(red)(cancel(color(black)(k))) + br(r-1) = color(red)(cancel(color(black)(k))) + b(r-1)#

Hence #b = 0# or #r = 1# (or both).

That means that the geometric sequence #b_0, b_1, b_2,...# must be constant.

Geometric sequence ?

If #b != 0# and #abs(r) > 1# then #c_(n+1)/c_n -> r# as #n -> oo#.

If #c_0, c_1, c_2,...# is a geometric sequence then the ratio between successive terms must be constant, so we require #c_1 / c_0 = r#

#r = c_1 / c_0 = (a+k+br)/(a+b)#

Multiply both ends by #(a+b)# to find:

#a+k+color(red)(cancel(color(black)(br))) = ar+color(red)(cancel(color(black)(br)))#

Also

#r = c_2 / c_1 = (a+2k+br^2)/(a+k+br)#

Multiply both ends by #(a+k+br)# to get:

#ar+kr+color(red)(cancel(color(black)(br^2))) = a+2k+color(red)(cancel(color(black)(br^2)))#

Hence we have #a+k = ar# and #a+2k = ar^2#

So:

#a^2 + 2ak = a(a+2k) = a^2r^2 = (ar)^2 = (a+k)^2#

#=a^2+2ak+k^2#

So #k = 0#

Then #ar = a + k = a#, so #a = 0# or #r = 1#

Well we assumed #abs(r) > 1# so #r != 1#, so #a = 0#

If #a=0# and #k=0# then the arithmetic sequence is constant #0#.

Now consider the case #abs(r) <= 1#.

If the arithmetic sequence is non-constant, then #c_(n+1)/c_n -> 1# as #n -> oo#. Hence we require the common ratio of the geometric sequence to be #1#. Then it's a constant sequence, therefore arithmetic and we require #b=0# or #r = 1# (or both). As a result we also require the arithmetic sequence to be constant, contradicting our supposition.