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
11 and 12 Grade Math
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