Answer with Explanation:
It is given that infinite geometric series with a beginning value of 2 , converges to 10.
Let a be the common ratio of geometric series.Where ,a<1.
Sum to infinity of a geometric series
\(\frac{a}{1-r} for, r<1 or \frac{a}{r-1}, for r>1.\)
\(2+2a+2a^2+2a^3+2a^4+.............{\text{to infinity}}=10\\\\ \frac{2}{1-a}=10\\\\ 1-a=\frac{2}{10}\\\\ 1-\frac{1}{5}=a\\\\a=\frac{4}{5}\)
First four terms of geometric series
\(2, 2*\frac{4}{5},2*(\frac{4}{5})^2,2*(\frac{4}{5})^3,2*(\frac{4}{5})^4.\\\\2, \frac{8}{5},\frac{32}{25},\frac{128}{125},\frac{512}{625}\)
