The sum of the first and third terms of a geometric sequence is 40 while the sum of its second and fourth terms is 96. How do you find the sixth term of the sequence?

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Answer 1 · 2016-01-15T19:20:04

Solve to find the initial term $a$ , common ratio $r$ and hence:

$a_6 = ar^5 = 1990656/4225$

The general term of a geometric sequence is:

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

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

We are given:

$40 = a_1 + a_3 = a + ar^2 = a(1+r^2)$

$96 = a_2 + a_4 = ar + ar^3 = ar(1+r^2)$

So:

$r = (ar(1+r^2))/(a(1+r^2)) = 96/40 = 12/5$

$a = 40/(1+r^2)$

$= 40/(1+(12/5)^2)$

$=40/(1+144/25)$

$=40/(169/25)$

$=(40*25)/169$

$=1000/169$

Then:

$a_6 = ar^5$

$= 1000/169*(12/5)^5$

$=(2^3*5^3*2^10*3^5)/(13^2*5^5)$

$=(2^13*3^5)/(5^2*13^2)$

$=1990656/4225$