Question #90e7d

1 Answer
Sep 22, 2016

Answer:

#a_n = 4*4^(n-1), n = 1, 2, 3, ...#

or
#a_n = 64*(1/4)^(n-1), n = 1, 2, 3, ...#

Explanation:

As this is posted under geometric sequences, we will proceed under the assumption that the sequences requested must be geometric. A geometric sequence is a sequence whose #n^"th"# term is given by #ar^(n-1)# where #a# is an initial value and #r# is a common ratio between terms.

By the above, the first #3# terms are #a, ar, ar^2#, and we are given the conditions that
#{(ar = 16), (a+ar+ar^2 = 84):}#

Substituting the first condition into the second and subtracting #16#, we can rewrite the second condition as

#a+ar^2 = 68# or #a(r^2+1)=68#

Dividing this by #ar = 16#, we get

#(a(r^2+1))/(ar) = 68/16#

#=> (r^2+1)/r = 17/4#

#=> r^2-17/4r + 1 = 0#

Applying the quadratic formula gives us

#r = (17/4+-sqrt((17/4)^2-4))/2 = (17+-15)/8#

So #r = 4# or #r = 1/4#

From #ar = 16#, we also have #a = 16/r#. Thus, by selecting the different possibilities for #r#, we get

#(a, r) = (4, 4)# or #(a, r) = (64, 1/4)#

Checking our answers, our first #3# terms for #(a, r) = (4, 4)# would be
#4, 16, 64#

and our first #3# terms for #(a, r) = (64, 1/4)# would be
#64, 16, 4#

Note that in each case, the sum of the terms is #84#, as desired.


As a side note, if we do not require a geometric sequence, any sequence #a_1, a_2, a_3, ...# would suffice so long as the second and third terms satisfy #a_2 = 16# and #a_3 = 68-a_1#.