The given G.P. is 0.15, 0.015, 0.00015, …
Here, a = 0.15 and r = \(\frac{0.015}{0.15}\) = 0.1
Sn = a\(\frac{(1-rn)}{1-r}\)
∴ S20 = 0.15 \(\frac{[1-(0.1)20]}{1-0.1}\)
= \(\frac{0.15}{0.9}[{1-(0.1)20]}\)
= \(\frac{15}{90}{[1-(0.1)20]}\)
= \(\frac{1}{6}{[1-(0.1)20]}\)
If the first and the nth term of a G.P. are a and b, respectively, and if P is the product of n terms, prove that P2 = (ab) n .