Today our topic of discussion is Geometric Series.
Geometric Series
Geometric Series
If the ratio of any term and its antecedent term of any series is always equal i.e., if any term divided by its antecedent term, the quotient is always equal, the series is called a geometric series and the quotient is called common ratio. Such as, of the series 2 + 4+8+ 16 + 32 the first term is 2, the second term is 4, the third term is 8, the fourth term is 16 and the fifth term is 32. Here,
8 the ratio of the third term to the second term = =2
the ratio of the second term to the first term 4 2 =2 = 2 16 2.
16 8 the ratio of the fourth term to the third term
the ratio of the fifth term to the fourth term 32
So, this series is a geometric series. In this series, the ratio of any term to its antecedent term is always equal. The common ratio of the mentioned series is 2. The numbers of terms of the series are finite. That is why the series is finite geometric series.
The geometric series is widely used in different areas of physical and biological science, in organizations like Banks and Life Insurance etc, and in different branches of technology.
If the numbers of terms are not fixed in a geometric series, it is called an infinite geometric series. The first term of a geometric series is generally expressed by a and common ratios by r. So by definition, if the first term is a, the second term is ar, the third term is ar² etc. Hence the series will be a + ar+ar² + ar³ +····
General term of a Geometric series
Let the first term of a geometric series be a and common ratio be r. Then, of the series
First term a ar¹-1
Second term = ar = a * r ^ (2 – 1)
Third term = ar² = ar³-1
Fourth term = a * r³ = a * r ^ (4 – 1)
nth term = ar”-1
This nth term is called the general term of the geometric series. If the first term of a geometric series a and the common ratio r are known, any term of the series can be determined by putting n = 1, 2, 3 ,*** etc. successively in the equation of the nth term.
Example 7. What is the 10th term of the series 2 + 4 + 8 + 16 +*** ?
Solution: The first term of the series a = 2 , common ratio r = 4/2 = 2
∴ The given series is a geometric series.
We know, the nth term of a geometric series = ar”-1
∴10th term of the series = 2 * 2 ^ (10 – 1) = 2 * 2 ^ 9 = 1024
Example 8. What is the general term of the series 128 + 64 + 32 +*** ?
Solution: The first term of the series a = 128 common ratio r = 64/128 = 1/2
∴It is a geometric series.
We know, general term of geometric series = ar”-1
So, the general term of the series
= 128 * (1/2) ^ (n – 1) = (2 ^ 7)/(2 ^ (n – 1)) = 1/(2 ^ (n – 1 – 7)) = 1/(2 ^ (n – 8))
Example 9. first and the second terms of a geometric series are 27 and 9. Find the 5th and the 10th terms of the series.
Solution: The first term of the given series a = 27 the second term = 9
∴The common ratio r = 9/27 = 1/3
∴ The fifth term = ar5-1 = 27 x 1 27 × 1 1 = 3 27 x 3 3
and the tenth term = a * r ^ (10 – 1)
= 27 * (1/3) ^ 9
= (3 ^ 3)/(3 ^ 3 * 3 ^ 6)
= 1/(3 ^ 6) = 1/729
Determination of the sum of a Geometric series
Let the first term of the geometric series be a, common ratio r and number of terms n. If Sn is the sum of n terms,
Sn = a + ar + a * r ² +***+ar^ n-2 +ar^ n-1 …(1)
and ** Sn =ar+ar² +ar^ 3 +***+ar^ n-1 + a * r ^ n * [multiplying(1) * byr] …(2)
Subtracting, Sn – r*Sn = a – a * r ^ n
or, Sn (1 – r) = a(1 – r ^ n)
Sn = (a(1 – r ^ n))/(1 – r) when r < 1
Again, subtracting (1) from (2),
r*Sn – Sn = a * r ^ n – a
or, Sn (r – 1) = a(r ^ n – 1)
Sn = (a(r ^ n – 1))/(r – 1) when r > 1
Observe: If common ratio is r = 1 , each term = a
Here, in this case S_{n} = a + a + a + upto n = an
Example 10. What is the sum of the series 12 + 24 + 48 +***+768?
Solution: The first term of the series is a = 12 , common ratio r = 24/12 = 2 > 1
∴ it is a geometric series.
Let, the nth term of the series = 768
We know, the nth term = ar”-1
a * r ^ (n – 1) = 768
or, 12 * 2 ^ (n – 1) = 768
or ,2^ n-1 = 768 12 =64
or, 2 ^ (n – 1) = 2 ^ 6
or, n – 1 = 6
∴ n = 7
Therefore, the sum of the series = (a(r – 1))/(r – 1) , when r > 1
= (12(2 ^ 7 – 1))/(2 – 1)
= 12(128 – 1)
= 12 * 127
= 1524
Example 11. Find the sum of the first eight terms of the series 1 + 1/2 + 1/4 + 1/8 +***
Solution: The first term of the series a = 1 , common ratio .. It is a geometric series. Here the number of terms n = 8 r = (1/2)/1 = 1/2 < 1
We know, sum of n terms of a geometric series S_{n} = (a(1 – r ^ n))/(1 – r) when r < 1
Hence, sum of eight terms of the series is
S8 = (1{1 – (1/2) ^ 8})/(1 – 1/2)
= (1 – 1/256)/(1/2)
= 2((256 – 1)/256)
= 255/128
= 1 127/128
Example 12. Palash Sarker joined the job in January 2005 at a yearly salary of Tk. 120000. His yearly increment is Tk. 5000. 10% of his salary is deducted every year for provident fund. At the end of year he deposits 12000 Tk. in a bank at a compound interest of 12%. He will retire from the job on December 31, 2030.
1) Which series done the basic salary of Palash Sarker follow? Write down that series.
2) How much salary (in Tk.) would he receive in his entire job life excluding the provident fund money?
3) What is the amount of total deposited money with interest in the bank by December 31, 2031?
Solution:
1) Basic salary of Palash Sarker follows arithmetic series. The first term of the series a = 120000 and common difference =5000
∴ The second term i = 120000 + 5000 = 125000
The third term = 125000+ 5000 130000
∴ The series is, 120000 + 125000 + 130000 +***
2) The total amount of salary excluding the provident fund from January 2005 to December 31, 2030 i.e. (2030-2005+ 1) or, 26 years is:
(120000-10% of 120000)+(125000-10% of 125000) + (130000 – 10% of 130000) + …
= (120000 – 12000) + (125000 – 12500) + (130000 – 13000) +***
= 108000 + 112500 + 117000 +***
In this cane it is an arithmetic series whose first term, a = 108000 common difference d = 112500 – 108000 = 4500 and number of terms n = 26
∴ Total salary he receives in 26 years = (2 x 108000+ (261) × 4500) 2 Tk.
= 13(216000 + 112500) = 13 * 328500 = 4270500Tl .
3) Total time from 2005 to 2031 is (2031-2005) or 26 years
Deposit of 12000 Tk. after 1 year 1200 (1 + 12/100) = 12000 * 1.12Tk
Deposit of 12000 Tk. after 2 years 12000 * (1.12) ^ 2 * Tk
Deposit of 12000 Tk. after 3 years 12000 * (1.12) ^ 3 * Tk
∴Total deposited amount after 26 years = 12000×1.12+12000×(1.12)²+… upto 26 th term.
=12000\ 1.12 + (1.12) ^ 2 +***+(1.12)^ 26
= 12000 * 1.12 * ((1.12) ^ 26 – 1)/(1.12 – 1) = 12000 * 1.12 * 18.04/0.12
=2020488 Tk. (approx.)
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