(-1 < r <1) or |r| < 1 is as follows

S = $\frac{a}{1 - r}$

Proof : Consider an infinite G.P with first as 'a' and common ratio as 'r' where |r|<1. We know that sum of 'n' terms of this G.P is given by

$S_{n} = a\left ( \frac{1-r^{n}}{1 - r} \right )$

⇒ $S_{n} =\frac{a}{1 - r} - \frac{ar^{n}}{1- r}$ -----------(1)

Since -1 < r < 1, so $r^{n}$ decreases as 'n' increase and $r^{n}$ tends to zero as 'n' tends to infinity. $r^{n}\rightarrow$ 0 as n $\rightarrow \infty $

∴ $\frac{ar^{n}}{1- r} \rightarrow$ 0 as n $\rightarrow \infty $

For |r| < 1

⇒ S = $\lim_{n\rightarrow \infty }S_{n} = \lim_{n\rightarrow \infty }\left ( \frac{a}{1 -r} - \frac{ar^{n}}{1 -r}\right )= \frac{a}{1 - r}$

common ratio = r = $\frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2}$ which is less than 1

S = $\frac{a}{1 - r}$

= $\frac{8}{1 - \frac{\sqrt{2}}{2}}$

S = $\frac{16}{2 - \sqrt{2}}$

Now rationalize the denominator

S = $\frac{16(2 + \sqrt{2}}{(2 - \sqrt{2})(2 + \sqrt{2}})$

S = $\frac{16(2 + \sqrt{2}}{(4 - 2)}$

S = $\frac{16(2 + \sqrt{2}}{2}$

∴ S = 8(2 + $\sqrt{2}$)

2) Finding the sum to infinity of the G.P 10 - 9 + 9.1 - 7.29 ...$\infty$

common ratio = r = $\frac{-9}{10}$ = -0.9 which is less than 1

S = $\frac{a}{1 - r}$

= $\frac{10}{1 - (-0.9)}$

S = $\frac{10}{1.9}$

∴ S = 5.263

3) If b = a + $a^{2} + a^{3}$ + ...$\infty$. Prove that a = $\frac{b}{1 + b}$

b = a + $a^{2} + a^{3}$ + ...$\infty$ From the above we can say that the above series is a Geometric series with first term 'a' and common ratio 'a'

S = $\frac{a}{1 - r}$

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

b(1 - a) = a

b - ab = a

∴ b = a + ab

∴ b = a (1 +b)

So, a = $\frac{b}{1 + b}$

From sum of an infinite geometric progression to Home

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