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Explainer Video

Tutor: Labib

Summary

Geometric series

​​In a nutshell

An geometric series finds the sum of a given number of terms in an geometric sequence, a sequence whose terms are separated by a common ratio. 



Calculating the sum of a sequence

To find the sum of a given number of terms of an geometric sequence, use a formula rather than manually adding up the terms. The sum SnS_nSn​ is given using the first term, aaa, the common ratio, rrr, and the nthn^{th}nth term.​

​Sn=a(1−rn)1−rorSn=a(rn−1)r−1\boxed {S_n = \dfrac {a(1-r^n)}{1-r} \quad or \quad S_n = \dfrac {a(r^n-1)}{r-1}}Sn​=1−ra(1−rn)​orSn​=r−1a(rn−1)​​​​


Note: Either formula can be used to find the sum but the formula on the left is easier when r<1r<1r<1 and the right when r>1r>1r>1.


Deriving the formula

The terms in a finite geometric sequence are as follows:

​a,ar1,ar2,...,arn−2,arn−1a, ar^1 ,ar^2,... , ar^{n-2}, ar^{n-1}a,ar1,ar2,...,arn−2,arn−1​​


This means the sum of a finite geometric sequence is equal to: 

​Sn=a+ar1+ar2+...+arn−2+arn−1S_n=a+ar^1 +ar^2+...+ ar^{n-2}+ ar^{n-1}Sn​=a+ar1+ar2+...+arn−2+arn−1​​


Multiplying the series by rrr will give: 

rSn=r[a+ar1+...+arn−2+arn−1]rSn=[ar+ar2+...+arn−1+arn]\begin{aligned} rS_n &=&r &[a+ar^1 +...+ ar^{n-2}+ ar^{n-1}] \\ rS_n &=& &[ar+ar^2 +...+ ar^{n-1}+ ar^{n}] \end{aligned}rSn​rSn​​==​r​[a+ar1+...+arn−2+arn−1][ar+ar2+...+arn−1+arn]​​​

​

Subtract this from SnS_nSn​:

Sn−rSn=[a+ar1+...+arn−2+arn−1]−[ar+ar2+...+arn−1+arn]S_n - rS_n = [a+ar^1 +...+ ar^{n-2}+ ar^{n-1}]- [ar+ar^2 +...+ ar^{n-1}+ ar^{n}]Sn​−rSn​=[a+ar1+...+arn−2+arn−1]−[ar+ar2+...+arn−1+arn]​​

​Sn(1−r)=a+[ar1−ar1]+...+[arn−2−arn−2]+[arn−1−arn−1]−arnS_n(1-r ) = a+[ar^1-ar^1] +...+ [ar^{n-2}-ar^{n-2}]+ [ar^{n-1}-ar^{n-1}]- ar^{n}Sn​(1−r)=a+[ar1−ar1]+...+[arn−2−arn−2]+[arn−1−arn−1]−arn​​

Sn(1−r)=a−arnS_n(1-r ) = a- ar^{n} Sn​(1−r)=a−arn


​Sn=a(1−rn)1−r\boxed{S_n= \dfrac{a(1-r^{n})}{1-r}}Sn​=1−ra(1−rn)​​


​Multiplying the top and bottom of the fraction by −1-1−1 which will maintain equivalence to SnS_nSn​:

Sn=a(1−rn)1−r×−1−1S_n= \dfrac{a(1-r^{n})}{1-r} \times \dfrac{-1}{-1} Sn​=1−ra(1−rn)​×−1−1​


Sn=a(rn−1)r−1\boxed{S_n= \dfrac{a(r^{n}-1)}{r-1}}Sn​=r−1a(rn−1)​​​​


Example 1

Find the sum of the first 121212 terms of the geometric sequence with a first term of, 777, and a common ratio of 0.90.90.9. 


Identify the formula to use:

Sn=a(1−rn)1−rS_n = \dfrac{a(1-r^n)}{1-r} Sn​=1−ra(1−rn)​​​

​

Substitute values to find the sum: 

S12=7(1−0.912)1−0.9S_{12} = \dfrac{7(1-0.9^{12})}{1-0.9} S12​=1−0.97(1−0.912)​​​

S12=50.23 (2 d.p.)‾\underline{S_{12} = 50.23\ (2 \ d.p.)}S12​=50.23 (2 d.p.)​​​


Example 2

How many terms are needed for the sum of a geometric series in the form 2+4+8+16+...2+4+8+16 +...2+4+8+16+... to exceed 100010001000?


Identify the common ratio and first term:

r=42=2a=2r=\dfrac 42 = 2 \\ a = 2r=24​=2a=2​​


Substitute the given values and solve:​

1000=2(2n−1)2−11000 = \dfrac{2(2^n-1)}{2- 1}1000=2−12(2n−1)​​​

​500=2n−1500= {2^n-1}500=2n−1​​

​501=2n501 = 2^n 501=2n

​ln⁡(501)=nln⁡(2)\ln (501) = n \ln(2)ln(501)=nln(2)​​

​n=ln⁡(501)ln⁡(2)=8.968666n = \dfrac{\ln(501)}{\ln(2)} = 8.968666n=ln(2)ln(501)​=8.968666


Round up since it is the sum which exceeds 100010001000:

​n=9‾\underline{n = 9} n=9​​​



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FAQs - Frequently Asked Questions

What is the formula for an geometric series?

S_n = a(1-r^n)/(1-r) or S_n = a(r^n-1)/(r-1) where a is the first term of a sequence, n is the position in a sequence, and r is the common ratio.

What is an geometric series based on?

A geometric sequence.

What is an geometric series?

An geometric series finds the sum of a given number of terms in an geometric sequence.

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