# 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)

____________________________________________________________

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

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