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

Let the common ratio of the G.P. be r.
Hence we have $$a_n= 3(a_{n+1}+ a_{n+2} + ...)$$

The sum up to infinity of GP is given by $$\frac{a}{1-r}$$ where a here is $$a_{n+1}$$

=> $$a_n= 3(\frac{a_{n+1}}{1-r})$$
=> $$a_n= 3(\frac{a_{n}\times r}{1-r})$$
=> $$ r = \frac{1}{4}$$
Now, $$a_1+a_2+a_2...+=32$$
=> $$\frac{a_1}{1-r} = 32$$
=> $$\frac{a_1}{3/4} = 32$$
=> $$a_1 = 24$$

$$a_5 = a_1 \times r^4$$
$$a_5 = 24 \times (1/4)^4 = \frac{3}{32}$$