How do you write a rule for the nth term of the geometric sequence, then find a6 given 2, 4/3, 8/9, 16/27,...?

1 Answer
May 31, 2016

Answer:

#T_6 = 64/243#

Explanation:

The first term is given as 2, so #a = 2#

The common ration, #r# is found by dividing 2 consecutive terms..
#r = (T_n)/(T_(n-1))# In other words, a term divided by the one before it.

#r = 4/3 ÷2/1 = 4/3 xx 1/2 = 2/3#

Check: #r = 8/9 ÷ 4/3 = 8/9 xx 3/4 =2/3#

Now, knowing the values for #a and r#, we can substitute into the general form for a term of a GP

#T_n = ar^(n-1) rArr 2 xx (2/3)^(n-1)#

Simplifying gives: # T_n =(2^n)/(3^(n-1)#

In the same way we can find #a_6 = ar^5#

But we have already found the general form, so let's use that for #n = 6#

#T_6 = (2^6)/(3^5)#

#T_6 = 64/243#