# Sum of an infinite Geometric Progression

The sum of an infinite Geometric Progression whose first term 'a' and common ratio 'r' (-1 < r < 1 i.e., |r| < 1) is

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

Proof:

A series of the form a + ar + ar$$^{2}$$ + ...... + ar$$^{n}$$ + ............... ∞ is called an infinite geometric series.

Let us consider an infinite Geometric Progression with first term a and common ratio r, where -1 < r < 1 i.e., |r| < 1. Therefore, the sum of n terms of this Geometric Progression in given by

S$$_{n}$$ = a($$\frac{1 - r^{n}}{1 - r}$$) = $$\frac{a}{1 - r}$$ - $$\frac{ar^{n}}{1 - r}$$ ........................ (i)

Since - 1< r < 1, therefore r$$^{n}$$  decreases as n increases and r^n tends to zero an n tends to infinity i.e., r$$^{n}$$ → 0 as n → ∞.

Therefore,

$$\frac{ar^{n}}{1 - r}$$ → 0 as n → ∞.

Hence, from (i), the sum of an infinite Geometric Progression ig given by

S = $$\lim_{x \to 0}$$ S$$_{n}$$  = $$\lim_{x \to \infty} (\frac{a}{ 1 - r} - \frac{ar^{2}}{1 - r})$$ = $$\frac{a}{1 - r}$$ if |r| < 1

Note: (i) If an infinite series has a sum, the series is said to be convergent. On the contrary, an infinite series is said to be divergent it it has no sum. The infinite geometric series a + ar + ar$$^{2}$$ + ...... + ar$$^{n}$$ + ............... ∞ has a sum when -1 < r < 1; so it is convergent when -1 < r < 1. But it is divergent when r > 1 or, r < -1.

(ii) If r ≥ 1, then the sum of an infinite Geometric Progression tens to infinity.

Solved examples to find the sum to infinity of the Geometric Progression:

1. Find the sum to infinity of the Geometric Progression

-$$\frac{5}{4}$$, $$\frac{5}{16}$$, -$$\frac{5}{64}$$, $$\frac{5}{256}$$, .........

Solution:

The given Geometric Progression is -$$\frac{5}{4}$$, $$\frac{5}{16}$$, -$$\frac{5}{64}$$, $$\frac{5}{256}$$, .........

It has first term a = -$$\frac{5}{4}$$ and the common ratio r = -$$\frac{1}{4}$$. Also, |r| < 1.

Therefore, the sum to infinity is given by

S = $$\frac{a}{1 - r}$$ = $$\frac{\frac{5}{4}}{1 - (-\frac{1}{4})}$$  = -1

2. Express the recurring decimals as rational number: $$3\dot{6}$$

Solution:

$$3\dot{6}$$ = 0.3636363636............... ∞

= 0.36 + 0.0036 + 0.000036 + 0.00000036 + .................. ∞

= $$\frac{36}{10^{2}}$$ + $$\frac{36}{10^{4}}$$ + $$\frac{36}{10^{6}}$$ + $$\frac{36}{10^{8}}$$ + .................. ∞, which is an infinite geometric series whose first term = $$\frac{36}{10^{2}}$$ and common ratio = $$\frac{1}{10^{2}}$$ < 1.

= $$\frac{\frac{36}{10^{2}}}{1 - \frac{1}{10^{2}}}$$, [Using the formula S = $$\frac{a}{1 - r}$$]

= $$\frac{\frac{36}{100}}{1 - \frac{1}{100}}$$

= $$\frac{\frac{36}{100}}{\frac{100 - 1}{100}}$$

= $$\frac{\frac{36}{100}}{\frac{99}{100}}$$

= $$\frac{36}{100}$$ × $$\frac{100}{99}$$

= $$\frac{4}{11}$$

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Geometric Progression