The second term of a geometric sequence is -18 and the fifth term is 2/3. What is the sixth term?

2 Answers
Feb 25, 2018

#a_6=-2/9#

Explanation:

#"using the nth term formula for a geometric sequence"#

#•color(white)(x)a_n=ar^(n-1)#

#rArra_2=ar=-18to(1)#

#rArra_5=ar^4=2/3to(2)#

#"divide equation "(2)" by equation "(1)#

#rArr(ar^4)/(ar)=(2/3)/(-18)=-1/27#

#rArrr^3=-1/27rArrr=root(3)(-1/27)=-1/3#

#"substitute this value in equation "(1)#

#axx-1/3=-18rArra=54#

#rArra_6=54xx(-1/3)^5=54xx-1/243=-2/9#

Feb 25, 2018

#-2/9#

Explanation:

The #n#th term of a geometric sequence is given by the formula:

#a_n = a r^(n-1)#

where #a# is the initial term and #r# the common ratio.

In our example, we find:

#-1/27 = (2/3)/(-18) = a_5/a_2 = (ar^4)/(ar) = r^3#

The only real solution of this is:

#r = root(3)(-1/27) = -1/3#

If we permit complex common ratios then there are two other possibilities, namely #-1/3omega# and #-1/3omega^2# where #omega = -1/2+sqrt(3)/2i# is the primitive complex cube root of #1#.

Assuming we want to use the real solution:

#a_6 = a_5 * r = -1/3(2/3) = -2/9#