Answer
The sum of given infinite geometric series = $\frac{80}{13}$
Work Step by Step
On put i = 1 to infinite in 8$\times$$(-0.3)^{i - 1}$, given summation can be written in the form of series as below
= 8$\times$$(-0.3)^{1 - 1}$ + 8$\times$$(-0.3)^{2 - 1}$ + 8$\times$$(-0.3)^{3 - 1}$ + 8$\times$$(-0.3)^{4 - 1}$ + ..................
= 8$\times$$(-0.3)^{0}$ + 8$\times$$(-0.3)^{1}$ + 8$\times$$(-0.3)^{2}$ + 8$\times$$(-0.3)^{3}$ ................
= 8 + 8$\times$$(-0.3)^{1}$ + 8$\times$$(-0.3)^{2}$ + 8$\times$$(-0.3)^{3}$ + ..................
Thus the infinite geometric series = 8 + 8$\times$$(-0.3)^{1}$ + 8$\times$$(-0.3)^{2}$ + 8$\times$$(-0.3)^{3}$ + ..................
Now first term $a_{1}$ = 8
Common ratio (r)= $\frac{8\times(-0.3)^{3}}{8\times(-0.3)^{2}}$
= $\frac{8\times(-0.3)^{2}}{8\times(-0.3)^{1}}$
= $\frac{8\times(-0.3)^{1}}{8}$
= -0.3
The sum of given infinite geometric series = $\frac{a_{1}}{1 - r}$
= $\frac{8}{1 - (-0.3)}$
= $\frac{8}{1 + 0.3}$
= $\frac{8}{1.3}$
= $\frac{80}{13}$
