How do you write a rule for the nth term of the geometric term given the two terms #a_4=-8/9, a_7=-64/243#?

1 Answer
Feb 28, 2017

Answer:

# a_n = -2^(n-1)/3^(n-2)#

Explanation:

Suppose using standard notation for a GP sequence that the first term is #a# and the common ratio is #r#, the first few terms of the sequence are:

# ( a, ar, ar^2, ar^3, ar^4 , ... } #

Assuming that the first term is #a_1#, then;

# a_1 = a #
# a_2 = ar #
# a_3 = ar^2 #
#vdots#
# a_n = ar^(n-1) #

Then #a_4=-8/9 \ \ \ => ar^3 = -8/9 \ \ \ \ \ \ .....[1] #
And, #a_7 = -64/243 => ar^6 = -64/243 \ \ \ .....[2] #

#Eq [2] divide Eq[1]# gives;

# \ \ (ar^6)/(ar^3) = (-64/243)/(-8/9) #
# :. r^3 = 64/243*9/8 #
# :. r^3 = 8/27 #
# :. \ \ r = 2/3 #

Subs #r^3 = 8/27# into #Eq[1]# gives:

# a*8/27 = -8/9 #
# :. a = -8/9*27/8 #
# \ \ \ \ \ \ \ = -3 #

And so the terms form a GP with #a=-3#, #r=2/3#

So the #n^(th)# term is given by:

# a_n = ar^(n-1) #
# \ \ \ \ = -3(2/3)^(n-1)#
# \ \ \ \ = -2^(n-1)/3^(n-2)#

Check

# n=4 => a_4 = -2^3/3^2 = -8/9#
# n=7 => a_7 = -2^6/3^5 = -64/243#