How do you write a rule for the nth term of the geometric sequence and then find #a_5# given #a_3=-28, r=7/3#?

2 Answers
Mar 16, 2017

Answer:

#-1372/9#

Explanation:

Let say the sequences are,
#a_1, a_2, a_3...a_n#, where,
#a_1 = a_1*r^0 =a_1*r^(1-1) #
#a_2 = a_1*r =a_1*r^(2-1) #
#a_3 = a_2*r = (a_1*r)*r = a_1*r^2 = (a_1*r)*r = a_1*r^(3-1)#
Therefore,
#T_n = a_1r^(n-1)#

given that #a_3 = -28 and r = 7/3#

#a_5 = a_1*r^(5-1) = a_1*r^4 = a_1*r^2*r^2#

#a_3=a_1*r^2# therefore,
# a_1*r^2*r^2 = a_3*r^2 = -28 * (7/3)^2 = -1372/9#

Mar 16, 2017

Answer:

#a_n=-36/7(7/3)^(n-1),a_5=-1372/9#

Explanation:

The standard geometric sequence is.

#a,ar,ar^2,ar^3,......,ar^(n-1)#

where a is the first term, r is the common ratio and the nth term is

#• ar^(n-1)#

Each term in the sequence is obtained by multiplying the previous term by r, the common ratio.

#a_3=ar^2=-28#

#rArraxx(7/3)^2=-28#

#rArraxx49/9=-28#

#rArra=-28xx9/49=-36/7#

#rArra_n=-36/7(7/3)^(n-1)larrcolor(red)" nth term formula"#

#rArra_5=-36/7xx(7/3)^4=-1372/9#