Infinite Geometric Series

Let the answer to the question be represented by $x$ :

• 1st Series:

$a/b + a/b^2 + a/b^3... = 4\equiv$

$\equiv 1/b (a + a/b + a/b^2...) = 4\equiv$

$\equiv 1/b (a + 4) = 4\equiv$

$\equiv a/b + 4/b = 4\equiv$

$\equiv a + 4 = 4b\equiv$

$\equiv a = 4b - 4$

• 2nd Series:

$a/(a + b) + a/(a + b)^2 + a/(a + b)^3... = x\equiv$

$\equiv 1/(a+b) (a + a/(a + b) + a/(a + b)^2...) = x\equiv$

$\equiv 1/(a+b) (a + x) = x\equiv$

$\equiv a + x = (a+b)x\equiv$

$\equiv a + x = ax + bx\equiv$

• ( Substitution of $a$ with $4b - 4$ )

$\equiv 4b - 4 + x = (4b - 4)x + bx\equiv$

$\equiv 4b - 4 + x = 4bx - 4x + bx\equiv$

$\equiv 5x - 5bx = 4 - 4b\equiv$

$\equiv 5x(1 - b) = 4(1 - b)\equiv$

$\equiv 5x = 4\equiv$

$\equiv\boxed{x = 0.8}$

Sum of infinite geometric progression is a/(1-r)... So we get in first case a/(b-1)=4 ie a/4=b-1 In second case, we get sum=a/(a+b-1)=4a/5a=0.8

$(a/b)(1+1/b+...)=(a/b)*(1/(1-1/b))=a/(b-1)=4$ $a=4(b-1)$ $(a/(a+b))(1/(1-1/(a+b)))=a/(a+b-1)$ $4(b-1)/(4(b-1)+b-1)$ $ans=4/5=\boxed{0.8}$

With geometric series one get by the first series: $\sum_{i=1}^\infty\frac a{b^i}= a\cdot\left(\sum_{i=0}^\infty\left(\frac1b\right)^i-1\right) = a\cdot\left(\frac{1}{1-\frac1b}-1\right)=\frac{a}{b-1}=4.$ So we get $a=4b-4$ .

Let $x$ denote the value of the second series. Apply the same arguments as for the first series and we get $x=\frac{a}{a+b-1}=\frac{4b-4}{4b-4+b-1}=\frac{4(b-1)}{5(b-1)}=\frac45=\boxed{0.8}.$

$\frac{a}{b}$ + $\frac{a}{b^2}$ + $\frac{a}{b^3}$ + ... = $\frac{a}{b-1}$

$\frac{a}{b-1}$ = 4

$\frac{a}{a+b}$ + $\frac{a}{(a+b)^2}$ + $\frac{a}{(a+b)^3}$ + ... = $\frac{a}{a+b-1}$ = $\frac{a}{a+b-1}$

For $\frac{a}{b-1}$ = 4, reciprocal and add 4 on both sides to get

$\frac{a+b-1}{a}$ = $\frac{5}{4}$

And finally reciprocal this to get the desired sum $\frac{a}{a+b-1}$ = $\frac{4}{5}$