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




11 and 12 Grade Math 

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