Question

if a=1+b+b^2+b^3+.........to infinite .where a<b<1.write b in terms of a​

Answer 1

Answer:

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

Step-by-step explanation:

We have given that

$a=1+b+b^{2}+b^{3}+......$ to infinite ...... (1)

And also given that a<b<1, i.e. b is a fraction less than one.

Now, the right-hand side is an example of an infinite G.P. series with first term 1 and the common ratio is b. So, we can sum it by using the formula of summation of infinite G.P. series.

The formula gives

$a+ar+ar^{2}+ar^{3}+..............$∞ = $\frac{a}{1-r}$  

{where -1 < r < 1}

So, we can write from equation (1),

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

⇒$\frac{1}{a}=1-b$

⇒$b=1-\frac{1}{a}$ (Answer)