Question #730a0

2 Answers
Mar 30, 2017

Answer:

#27#

Explanation:

Denote the #n#:th element in the sequence #a_n = bx^(n-1)#.
Then #a_3=bx^2# and #a_6 = bx^5#. Knowing that #a_3 = 3# and #a_6=1/9#, we get that
#3 = b*x^2#
#1/9 = b*x^5 = (bx^2)x^3#
Substituting the first line into the second we get that
#1/9 = 3x^3#, which gives that
#x = 1/3#.
Inserting into the first equation, we get that
#b = 3/(x^2) = 27#.
Thus the first element #a_1 = bx^(1-1) = b = 27#.

Mar 30, 2017

Answer:

#27.#

Explanation:

Let the Geom. Seq. be #a,ar, ar^2,...,ar^(n-1),...#

By what is given, #ar^2=3, and, ar^5=1/9.#

#:. (ar^5)/(ar^2)=1/9-:3#.

#:. r^3=1/27#

#:. r=1/3.#

Now, #ar^2=3, r=1/3 rArr a=3-:r^2=3-:(1/3)^2=3-:1/9=27.#

#:." The First term is "27.#