a(n) = a(1)*r^(n-1), where a(1) is the first term, n is the counter and r is the common ratio.
I first noted that 243 is a power of 3; specifically, 243=3^5, or 243=3(3)^4, or 243=(3^2)(3)^(4-1). Notice that I'm trying here to rewrite 243=3^5 in the form a(n) = a(1)*r^(n-1): a(4) = a(1)(3)^(4-1), or a(4) = a(1)(3)^3 = 243. Then by division we find that a(1) = 243/27 = 9. Is it possible that a(1)=9?
Let's try out our formula a(n)=9(3)^(n-1). Steal n=9 and see whether this formula gives u s 59049:
n(9) = 59049 = 9(3)^(9-1), or 9(3)^8. True or false? 3^8= 6561, and 9(3)^8 = 59049.
YES! That's correct.
Therefore, the desired formula is
a(n) = 9(3)^(n-1). The first term, a(1) is 9(3)^(1-1) = 9(3)^0 = 9*1 = 9.
