Answer:
This is a geometric progresion that begins with 1 and each term is 1/3 the preceeding term
Let Pn represent the nth term in the sequence
Then Pn = (1/3)^n-1
From this P14 = (1/3)^13 = 1/1594323
5. The sum of the first n terms of a GP beginning a with ratio r is given by
Sn = a* (r^n+1 - 1)/(r - 1)
With n = 10, a = 1 and r = 1/3, S10 = ((1/3)^11 - 1)/(1/3 - 1) = 1.500
3/5*10 is the same as 3/5*10/1 and it should be 30/5 which is the same as 6.
So, 3/5*10 is 6
1 and 16, 2 and 8, 4 and 4.
