An infinite geometric progression $a_1, a_2, a_3,...$ has the property that $a_n = 3(a_{n+1} + a_{n+2} +....)$ for every $n \ge 1$. If the sum $a_1 + a_2 + a_3 +..... = 32$, then $a_5$ is

We have an infinite geometric progression where each term equals 3 times the sum of all terms that come after it.

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Given: $a_n = 3(a_{n+1} + a_{n+2} + a_{n+3} + ...)$ for every $n \geq 1$

For the first term: $a_1 = 3(a_2 + a_3 + a_4 + ...)$

In a geometric progression, if the first term is $a$ and common ratio is $r$, then:

$a_1 = a$

$a_2 = ar$

$a_3 = ar^2$

$a_4 = ar^3$, and so on...

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The sum $a_2 + a_3 + a_4 + ...$ is also a geometric series starting from $a_2 = ar$.

For infinite geometric series where $|r| < 1$: $\text{sum} = \dfrac{\text{first term}}{1 - r}$

Therefore: $a_2 + a_3 + a_4 + ... = \dfrac{ar}{1-r}$

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Substituting into our main equation:

$a_1 = 3(a_2 + a_3 + a_4 + ...)$

$a = 3 \cdot \dfrac{ar}{1-r}$

Dividing both sides by $a$:

$1 = \dfrac{3r}{1-r}$

$1-r = 3r$

$1 = 4r$

$r = \dfrac{1}{4}$

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Given that the total sum is 32:

$a_1 + a_2 + a_3 + ... = 32$

Using the infinite series formula:

$\dfrac{a_1}{1-r} = 32$

$\dfrac{a_1}{1-\frac{1}{4}} = 32$

$\dfrac{a_1}{\frac{3}{4}} = 32$

$a_1 = 32 \times \dfrac{3}{4} = 24$

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Now we can find each term:

$a_1 = 24$

$a_2 = 24 \times \dfrac{1}{4} = 6$

$a_3 = 6 \times \dfrac{1}{4} = \dfrac{6}{4} = \dfrac{3}{2}$

$a_4 = \dfrac{3}{2} \times \dfrac{1}{4} = \dfrac{3}{8}$

$a_5 = \dfrac{3}{8} \times \dfrac{1}{4} = \dfrac{3}{32}$

Alternative method: $a_5 = a_1 \times r^4 = 24 \times \left(\dfrac{1}{4}\right)^4 = 24 \times \dfrac{1}{256} = \dfrac{24}{256} = \dfrac{3}{32}$

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Therefore, $a_5 = \dfrac{3}{32}$

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When dealing with infinite geometric progressions, the sum formula only works when $|r| < 1$, and each term relationship can help us find the common ratio.