A sequence of number where each term is formed by multiplying a constant to the previous term.

      \(S_n=\dfrac{a(1-r^n)}{1-r}\)     

\(S_n=\dfrac{a(r^n-1)}{r-1}\)

where
\(r\neq 1\)

      \(S_{\infty}=\dfrac{a}{1-r}\)     

Note that

\(\begin{array}{ll} ar^3=24 && (1)\\ ar^5=10\dfrac{2}{3}=\dfrac{32}{3} && (2) \end{array}\)
 

\((2)\div (1):\\ \begin{aligned} \dfrac{ar^5}{ar^3}&= \dfrac{32}{3}\times \dfrac{1}{24}\\ r^2&=\dfrac{4}{9}\\ r&= \pm \dfrac{2}{3}. \end{aligned}\)

Since all the terms are positive,
then \(r=\dfrac{2}{3}\).

\(\begin{aligned} a\times \left(\dfrac{2}{3}\right)^3&=24\\ a \times \dfrac{8}{27}&=24\\ a&= 24\times \dfrac{27}{8}\\ a&=81. \end{aligned}\)

Therefore,

\(T_8=81\left(\dfrac{2}{3}\right)^7=\dfrac{128}{27}.\)

Note that

\(a=18\) and
\(r=\dfrac{6}{18}=\dfrac{1}{3}\).
 

\(\begin{aligned}T_n &\lt 0.0003\\ 18 \times \left(\dfrac{1}{3}\right)^{n-1} &\lt 0.0003\\ \left(\dfrac{1}{3}\right)^{n-1} &\lt \dfrac{0.0003}{18}\\ \log \left(\dfrac{1}{3}\right)^{n-1} &\lt \log \dfrac{0.0003}{18}\\ (n-1)\log \left(\dfrac{1}{3}\right) &\lt \log \dfrac{0.0003}{18}\\ n-1 &\gt \cfrac{\log \cfrac{0.0003}{18}}{\log \cfrac{1}{3}}\\ n-1 &\gt 10.01\\ n &\gt 11.01. \end{aligned}\)

\(\therefore n=12.\)

Therefore,

\(T_{12}=18\left(\dfrac{1}{3}\right)^{11}=0.0001016.\)

\(0.7777... \\= 0.7+0.07+0.007+...\)

Note that
\(a=0.7\) and
\(r=\dfrac{0.07}{0.7}= 0.1\).

Therefore,

\(\begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\&=\dfrac{0.7}{1-0.1}\\&= \dfrac{0.7}{0.9}\\&= \dfrac{7}{9}. \end{aligned}\)

\(0.151515...\\=0.15+0.0015+0.000015...\)

Note that
\(a=0.15\) and
\(r=\dfrac{0.0015}{0.15}= 0.01\).

Therefore,

\(\begin{aligned}S_{\infty}&=\dfrac{a}{1-r}\\&=\dfrac{0.15}{1-0.01}\\&= \dfrac{0.15}{0.99}\\&= \dfrac{15}{99}\\ &= \dfrac{5}{33}. \end{aligned}\)