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 \(\frac{[1-(0.1)20]}{1-0.1}\)
= \(\frac{0.15}{0.9}[{1-(0.1)20]}\)
= \(\frac{15}{90}{[1-(0.1)20]}\)
= \(\frac{1}{6}{[1-(0.1)20]}\)
\(f(x) = \begin{cases} x^2, & \quad 0≤x≤3\\ 3x, & \quad 3≤x≤10 \end{cases}\)
The relation g is defined by
\(g(x) = \begin{cases} x^2, & \quad 0≤x≤2\\ 3x, & \quad 2≤x≤10 \end{cases}\)
Show that f is a function and g is not a function.