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

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

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

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

= $\frac{36}{100}$ × $\frac{100}{99}$

= $\frac{4}{11}$

● Geometric Progression

• Definition of Geometric Progression
• General Form and General Term of a Geometric Progression
• Sum of n terms of a Geometric Progression
• Definition of Geometric Mean
• Position of a term in a Geometric Progression
• Selection of Terms in Geometric Progression
• Sum of an infinite Geometric Progression
• Geometric Progression Formulae
• Properties of Geometric Progression
• Relation between Arithmetic Means and Geometric Means
• Problems on Geometric Progression

11 and 12 Grade Math 

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