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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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 $a$ $a$ , common ratio $r$ $r$ and hence:

$a_{6} = a r^{5} = \frac{1990656}{4225}$ $a_6 = ar^5 = 1990656/4225$

Explanation:

The general term of a geometric sequence is:

$a_{n} = a r^{n - 1}$ $a_n = a r^(n-1)$

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

We are given:

$40 = a_{1} + a_{3} = a + a r^{2} = a (1 + r^{2})$ $40 = a_1 + a_3 = a + ar^2 = a(1+r^2)$

$96 = a_{2} + a_{4} = a r + a r^{3} = a r (1 + r^{2})$ $96 = a_2 + a_4 = ar + ar^3 = ar(1+r^2)$

So:

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

$a = \frac{40}{1 + r^{2}}$ $a = 40/(1+r^2)$

$= \frac{40}{1 + {(\frac{12}{5})}^{2}}$ $= 40/(1+(12/5)^2)$

$= \frac{40}{1 + \frac{144}{25}}$ $=40/(1+144/25)$

$= \frac{40}{\frac{169}{25}}$ $=40/(169/25)$

$= \frac{40 \cdot 25}{169}$ $=(40*25)/169$

$= \frac{1000}{169}$ $=1000/169$

Then:

$a_{6} = a r^{5}$ $a_6 = ar^5$

$= \frac{1000}{169} \cdot {(\frac{12}{5})}^{5}$ $= 1000/169*(12/5)^5$

$= \frac{2^{3} \cdot 5^{3} \cdot 2^{10} \cdot 3^{5}}{13^{2} \cdot 5^{5}}$ $=(2^3*5^3*2^10*3^5)/(13^2*5^5)$

$= \frac{2^{13} \cdot 3^{5}}{5^{2} \cdot 13^{2}}$ $=(2^13*3^5)/(5^2*13^2)$

$= \frac{1990656}{4225}$ $=1990656/4225$

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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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Explanation:

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