Sum of n terms of a Geometric Progression

We will learn how to find the sum of n terms of the Geometric Progression {a, ar, ar$^{2}$, ar$^{3}$, ar$^{4}$, ...........}

To prove that the sum of first n terms of the Geometric Progression whose first term ‘a’ and common ratio ‘r’ is given by

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

⇒ S$_{n}$ = a($\frac{1 - r^{n}}{1 - r}$), r ≠ 1.

Let Sn denote the sum of n terms of the Geometric Progression {a, ar, ar$^{2}$, ar$^{3}$, ar$^{4}$, ...........} with first term ‘a’ and common ratio r. Then,

Now, the nth terms of the given Geometric Progression = a ∙ r$^{n - 1}$.

Therefore, S$_{n}$ = a + ar + ar$^{2}$ + ar$^{3}$ + ar$^{4}$ + ............... + ar$^{n - 2}$ + ar$^{n - 1}$ ............ (i)

Multiplying both sides by r, we get,

rS$_{n}$ = ar + ar$^{2}$ + ar$^{3}$ + ar$^{4}$ + ar$^{4}$ + ................ + ar$^{n - 1}$ + ar$^{n}$ ............ (ii)

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On subtracting (ii) from (i), we get

S$_{n}$ - rS$_{n}$ = a - ar$^{n}$

⇒ S$_{n}$(1 - r) = a(1 - r$^{n}$)

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

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

Hence, S$_{n}$ = a$\frac{(1 - r^{n})}{(1 - r)}$ or, S$_{n}$ = a$\frac{(r^{n} - 1)}{(r - 1)}$

Notes:

(i) The above formulas do not hold for r = 1. For r = 1, the sum of n terms of the Geometric Progression is S$_{n}$ = na.

(ii)When the numerical value of r is less than 1 (i.e., - 1 < r < 1), then the formula S$_{n}$ = a$\frac{(1 - r^{n})}{(1 - r)}$ is used.

(iii) When the numerical value of r is greater than 1 (i.e., r > 1 or, r < -1) then the formula S$_{n}$ = a$\frac{(r^{n} - 1)}{(r - 1)}$ is used.

(iv) When r = 1, then S$_{n}$ = a + a + a + a + a + .................... to n terms = na.

(v) If l is the last term of the Geometric Progression, then l = ar$^{n - 1}$.

Therefore, S$_{n}$ = a($\frac{1 - r^{n}}{1 - r}$) = ($\frac{a - ar^{n}}{1 - r}$) = $\frac{a - (ar^{n - 1})r}{(1 - r)}$ = $\frac{a - lr}{1 - r}$

Thus, S$_{n}$ = $\frac{a - lr}{1 - r}$

Or, S$_{n}$ = $\frac{lr - a}{r - 1}$, r ≠ 1.

Solved examples to find the Sum of first n terms of the Geometric Progression:

1. Find the sum of the geometric series:

4 - 12 + 36 - 108 + .............. to 10 terms

Solution:

The first term of the given Geometric Progression = a = 4 and its common ratio = r = $\frac{-12}{4}$ = -3.

Therefore, the sum of the first 10 terms of the geometric series

= a ∙ $\frac{r^{n} - 1}{r - 1}$, [Using the formula S$_{n}$ = a$\frac{(r^{n} - 1)}{(r - 1)}$ since, r = - 3 i.e., r < -1]

= 4 ∙ $\frac{(-3)^{10} - 1}{-3 - 1}$

= 4 ∙ $\frac{(-3)^{10} - 1}{-4}$

= - (-3)$^{10}$ - 1

= -59048

2. Find the sum of the geometric series:

1 + $\frac{1}{2}$ + $\frac{1}{4}$ + $\frac{1}{8}$ + $\frac{1}{16}$ + .............. to 10 terms

Solution:

The first term of the given Geometric Progression = a = 1 and its common ratio = r = $\frac{\frac{1}{2}}{1}$ = $\frac{1}{2}$

Therefore, the sum of the first 10 terms of the geometric series

S$_{10}$ = a$\frac{(1 - r^{10})}{(1 - r)}$

⇒ S$_{10}$ = 1 ∙ $\frac{(1 - (\frac{1}{2})^{10})}{(1 - \frac{1}{2})}$

⇒ S$_{10}$ = 2($\frac{1}{2^{10}}$)

⇒ S$_{10}$ = 2($\frac{2^{10} - 1}{2^{10}}$)

⇒ S$_{10}$ = 2($\frac{1024 - 1}{1024}$)

⇒ S$_{10}$ = $\frac{1024 - 1}{512}$

⇒ S$_{10}$ = $\frac{1023}{512}$

Note that we have used formula Sn = a($\frac{(1 - r^{n})}{(1 - r)}$ since r = 1/4 i.e., r < 1]

3. Find the sum of 12 terms of the Geometric Progression 3, 12, 48, 192, 768, ................

Solution:

The first term of the given Geometric Progression = a = 3 and its common ratio = r = $\frac{12}{3}$ = 4

Therefore, the sum of the first 12 terms of the geometric series

Therefore, S$_{12}$ = a$\frac{r^{12} - 1}{r - 1}$

= 3($\frac{4^{12} - 1}{4 - 1}$)

= 3($\frac{16777216 - 1}{3}$)

= 16777216 - 1

= 16777215

4. Find the sum to n terms: 5 + 55 + 555 + 5555 + .............

Solution:

We have 5 + 55 + 555 + 5555 + ............. to n terms

= 5[1 + 11 + 111 + 1111 + .............. + to n terms]

= $\frac{5}{9}$[9 + 99 + 999 + 9999 + ................ + to n terms]

= $\frac{5}{9}$[(10 – 1) + (10$^{2}$ - 1) + (10$^{3}$ - 1) + (10$^{4}$ - 1) + .............. + (10$^{n}$ - 1)]

= $\frac{5}{9}$[(10 + 10$^{2}$ + 10$^{3}$ + 10$^{4}$ + ................ + 10$^{n}$) – ( 1 + 1 + 1 + 1 + ................ + 1)] n times

= $\frac{5}{9}$[10 × $\frac{(10^{n} - 1)}{(10 - 1)}$ – n]

= $\frac{5}{9}$[$\frac{10}{9}$(10$^{n}$ – 1) – n]

= $\frac{5}{81}$[10$^{n + 1}$ – 10 – 9n]

● 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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