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Mathematics Dictionary
Dr. K. G. Shih
Geometric Progression
Subjects

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  • Q01 | - What is geometric sereis ?
  • Q02 | - Formula of geometric series
  • Q03 | - Find sum of series 1 + 1/2 + 1/4 + 1/8 + 1/16 + ..... + n
  • Q04 | - Geometric means
  • Q05 | - (1-x^2)/(1-x) = 1 + x + x^2 + x^3 + x^4 + ....
  • Q06 | - Infinite geometric series of 0.3, + 0.03 + 0.003 + 0.0003 + ....
  • Q07 | - Each year saves $1000 with interest rate 10% for 20 years. Find saving amount
  • Q08 | - In G.P. we know 2nd term equal 1st tem less 1. Also sum of first 3 terms = 19/3
  • Q09 | - Repeating decimal 1.21212121 to rational number
  • Q10 | - Compound annually rate
  • Q11 | - Formula and outlines
Answers


Q01. What is arithmetic series ?
  • A series a1 + a2 + a3 + ...... + an
  • is called a geometric seires if, and only if, the ratios of succesive terms are all equal, that is
  • a2/a1 = a3/a2 = a4/a3 = ........
  • The ratio is call common ratio.
  • The series is given as a + a*r + a*r^2 + a*r^3 + .... a*r^(n-1)

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Q02. Formula
The terms
  • Let first term be T(1) = a and common ratio is r.
  • Then 2nd term is T(2) = a*r^1
  • Then 3rd term is T(3) = a*r^2
  • Then 4th term is T(4) = a*r^3
  • Hence the nth term is T(n) = a*r^(n-1)
The sum
  • Let the sum of n terms be S(n)
  • Then S(n) = Sum[T(n)] = a*(1-r^n)/(1-r).
  • If r is less than 1 and greater than -1,
    • then S(n) = a/(1-r)
    • Then the series is converge. That the sum will be fixed constant
Prove the formula
  • S(n) = a + a*r + a*r^2 + a*r^3 + ..... a*r^(n-1) ............. (1)
  • r*S(n) = + a*r + a*r^2 + a*r^3 + ..... a*r^(n-1) + a*r^n ..... (2)
  • (2) - (1) we have
  • r*S(n) - S(n) = a*r^n - a
  • Hence (r-1)*S(n) = a*(r^n-1)
  • and S(n) = a*(r^n - 1)/(r - 1)

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Q03. Find sum of series 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/(r^n)
  • This is G.P.
  • Hence a = 1 and r = 1/2.
  • Use formula we have S(n) = a*(r^n - 1)/(r - 1)
  • Hence we have Sum(n) = (1)*((1/2)^n - 1)/(1/2 - 1).
  • S(n) = 2*(1 - (1/2)^n)

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Q04. Geometric means
  • Definition
    • In geometric series between the 1st and last term are called geometric mean.
  • If a,b,c are 3 consecutive terms in an G.P. then geometric mean is b^2 = a*c.
    • Since a,b,c are consecutive terms in geometric series
    • Hence a/b = b/c
    • Hence b^2 = a*c

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Q05. (1-x^n)/(1-x) = 1 + x + x^2 + x^3 + x^4 + .... + x^n
  • Right hand side is G.P.
  • a = 1 and r = x
  • Hence S = a*(r^n - 1)/(r - 1) = (1 - x^n)/(1 - x)

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Q06. Infinite geometric series of 0.3, + 0.03 + 0.003 + 0.0003 + ....
Prove that the sum is 1/3
  • 3*(0.1 + 0.01 + 0.001 + 0.0001 + ...)
  • S = 0.1 + 0.01 + 0.001 + 0.0001 + ..... is G.P. with a = 0.1 and r = 0.1
  • Hence S = a/(1 - r) = 0.1/(0.9) = 1/9
  • Hence the sum = 3*S = 1/3
  • This is to change repeating decimal 0.33333.... = 1/3
Other method
  • Let x = 0.333333....
  • Hence 10*x = 3.333333....
  • Hence 9*x = 3
  • Hence x = 3/9 = 1/3

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Q07. Each year saves $1000 with interest rate 10% for 20 years. Find saving amount
Formula
  • At end of 1st year the amount A1 = p*(1+i)
  • At end of 2nd year : A2 = p*(1+i) + A1*(1+i) = p*(1+i) + p*(1+i)^2
  • At end of 3nd year : A3 = p*(1+i) + A2*(1+i) = p*(1+i) + p*(1+i)^2 +p*(1+i)^3
  • This is G.P. with 1st term a = p*(1+i) and ratio = (1+i).
  • Since p = 1000, i =0.1 hence a = 1000*(1.1).
  • Ratio r = 1.1 and n = 20
  • Sum of seris : S = a*(r^n - 1)/(r - 1) = 1100*((1.1)^20 - 1)/(1.1 - 1)
  • S = 1100*(6.7274999 - 1)/(0.1) = 63002.50

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Q08. In G.P. we know 2nd term equal 1st tem less 1. Also sum of first 3 terms = 19/3 Find a and r
  • Let first term be a and ratio be r.
  • Hence a*r = a - 1 .................... (1)
  • Hence a + a*r + a*r^2 = 19/3 ......... (2)
  • From (1) we have a = 1/(1-r) ......... (3)
  • Substitue (3) into (2) : (1/(1-r))*(1 + r + r^2) = 19/3
  • Hence 3*(1 + r + r^2) = 19*(1 - r)
  • 3*r^2 + 3*r + 19*r + 3 - 19 = 0
  • 3*r^2 + 22*r - 16 = 0
  • r = (-22 + Sqr(22^2 - 4*(3)*(-16))/(2*3) = (-22 + 26)/6 = 4/6 = 2/3
  • From (3) we have a = 1/(1 - r) = 3
  • Verify
    • 1st term = 3
    • 2nd term = 2
    • 3rd term = a*r^2 = 3*(2/3)^2 = 4/3
    • Sum of first 3 terms = 3 + 2 + 4/3 = 19/3.
  • r = (-22 - Sqr(22^2 - 4*(3)*(-16))/(2*3) = (-22 - 26)/6 = -8
  • from (3) a = 1/(1+8) = 1/9.
  • Verify
    • 1st term = 1/9
    • 2nd term = 1/9 - 1 = -8/9
    • 3rd term = a*r^2 = (1/9)*(64)
    • Sum = 1/9 - 8/9 + 64/9 = (1 - 8 + 64)/9 = 57/9 = 19/3

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Q09. Repeating decimal 1.21212121 to rational number

G.P. method
  • 1.212121 = 1 + 0.21 + 0.0021 + 0.000021 + ....
  • S = 0.21 + 0.0021 + 0.000021 + .... is A.P. and a = 0.21 and r = 0.01
  • S = a/(1-r) = 0.21/(1-0.01) = 21/99 = 7/33
  • Hence the answer is 1 + S = 1 + 7/33 = 40/33.
Special method
  • Let x = 1.212121.....
  • 100*x = 121.212121...
  • 100*x - x = 121.212121... - 1.212121... = 120.
  • Hence 99*x = 120 and x = 120/99 = 40/33.

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Q10. Save $10000 in bank with annual compound rate 6% for 10 years
  • For the end of 1st year : A1 = P*(1 + i)
  • For the end of 2nd year : A2 = A1*(1 + i) = p*(1 + i)^2
  • For the end of 3rd year : A3 = A2*(1 + i) = p*(1 + i)^3
  • For the end of n year will be A = p*(1 + i)^n
  • Now p = 10000, i = 6% = 0.06, n = 10
  • The total amount is A = 10000*(1 + 0.06)^10 = 10000*(1.06)^10 = 17908.48

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Q11. Formula and outlines
Formula for G.P.
  • The nth term : T(n) = a*(r^(n-1))
  • The sum of n terms : Sum[T(n)] = a*(r^n - 1)/(r - 1)
  • If a,b,c are consecutive G.P. terms, then b^2 = a*c
Application
  • Repeating decimal to rational number
  • Bank interest calculation

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