Answer
The sum of given infinite geometric series = $\frac{120}{17}$
Work Step by Step
On put i = 1 to infinite in 12$\times$$(-0.7)^{i - 1}$, given summation can be written in the form of series as below
= 12$\times$$(-0.7)^{1 - 1}$ + 12$\times$$(-0.7)^{2 - 1}$ + 12$\times$$(-0.7)^{3 - 1}$ + 12$\times$$(-0.7)^{4 - 1}$ + ..................
= 12$\times$$(-0.7)^{0}$ + 12$\times$$(-0.7)^{1}$ + 12$\times$$(-0.7)^{2}$ + 12$\times$$(-0.7)^{3}$ ................
= 12 + 12$\times$$(-0.7)^{1}$ + 12$\times$$(-0.7)^{2}$ + 12$\times$$(-0.7)^{3}$ + ..................
Thus the infinite geometric series = 12 + 12$\times$$(-0.7)^{1}$ + 12$\times$$(-0.7)^{2}$ + 12$\times$$(-0.7)^{3}$ + ..................
Now first term $a_{1}$ = 12
Common ratio (r)= $\frac{12\times(-0.7)^{3}}{12\times(-0.7)^{2}}$
= $\frac{12\times(-0.7)^{2}}{12\times(-0.7)^{1}}$
= $\frac{12\times(-0.7)^{1}}{12}$
= -0.7
The sum of given infinite geometric series = $\frac{a_{1}}{1 - r}$
= $\frac{12}{1 - (-0.7)}$
= $\frac{12}{1 + 0.7}$
= $\frac{12}{1.7}$
= $\frac{120}{17}$
