Lesson Geometric progressions

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Geometric progressions


Definition
A  geometric progression  is a sequence of numbers such that the ratio of the current term to the preceding term is the same for any two consecutive terms.
The constant value of the ratio of the current term to the preceding term of the geometric progression is called the  common ratio  of the geometric progression


Examples


1. The sequence of the numbers

    1, 2, 4, 8, 16, 32

is the geometric progression because the ratio of the current term to the preceding term is the constant value equal to  2  for any two consecutive terms.
The first term of this progression is equal to  1,  the common ratio is equal to  2.
We can present this progression as a sequence: a%5B1%5D=1, a%5B2%5D=2, a%5B3%5D=4, a%5B4%5D=8, a%5B5%5D=16, a%5B6%5D=32.  Below we will use an index n to designate the current term
of a sequence.  For instance, if n=3,  a%5Bn%5D+=+a%5B3%5D+=+4  is the third term of the progression.


2. The sequence of the numbers

1, 1%2F2, 1%2F4, 1%2F8, 1%2F16, 1%2F32

is the geometric progression because the ratio of the current term to the preceding term is the constant value equal to  1%2F2  for any two consecutive terms.
The first term of this progression is equal to  1,  the common ratio is equal to  1%2F2.


3. The sequence of the numbers

    1, 10, 100, 1000, 10000, 100000

is the geometric progression.  The ratio of the current term to the preceding term is the constant value equal to 10 for any two consecutive terms.
The first term of this progression is equal to  1,  the common ratio is equal to  10.  This progression is an increasing sequence: the next term is bigger than
the previous one for any two consecutive terms.


4. The sequence of the numbers

    1, 0.1, 0.01, 0.001, 0.0001, 0.00001

is the geometric progression.  The ratio of the current term to the preceding term is the constant equal to  1%2F10  for any two consecutive terms.
The first term of this progression is equal to  1,  the common ratio is equal to  1%2F10,  or 0.1.  This progression is a decreasing sequence: the next
term is lesser than the previous one for any two consecutive terms.


5. The sequence of the numbers

1%2F3,  -1%2F9,  1%2F27,  -1%2F81

is the geometric progression.  The ratio of the current term to the preceding term is the constant value equal to  -1%2F3  for any two consecutive terms.
The first term of this progression is equal to  1%2F3,  the common ratio is equal to -1%2F3.


6. The sequence of the numbers

 1,  sqrt%282%29,  2,  2sqrt%282%29,  4,  4sqrt%282%29

is the geometric progression.  The ratio of the current term to the preceding term is the constant value equal to  sqrt%282%29  for any two consecutive terms.
The first term of this progression is equal to  1,  the common ratio is equal to  sqrt%282%29.


Formula for n-th term of a geometric progression

If  a%5B1%5D, a%5B2%5D, a%5B3%5D, . . . , a%5Bn%5D  is the geometric progression with the first term  a%5B1%5D  and the common ratio  q then the formula for the n-th term is

      a%5Bn%5D = a%5B1%5D%2Aq%5E%28n-1%29.                                                  (1)


For the proof of the formula (1) see the lesson
    The proofs of the formulas for geometric progressions
under the current topic in this site.


Examples


7. Find the  12-th term of the geometric progression if the first term is  3  and the common ratio is  2.

Solution.
Apply the formula  (1)  for the n-th term of a geometric progression.  You have

a%5B12%5D = a%5B1%5D%2Aq%5E%28n-1%29 = 3%2A2%5E%2812-1%29 = 3%2A2%5E11 = 3%2A2048 = 6144.



8. Find the 18-th term of the geometric progression if the first term is  5  and the common ratio is  sqrt%282%29.

Solution.
Apply the formula  (1)  for the n-th term of a geometric progression.  You have

a%5B18%5D = a%5B1%5D%2Aq%5E%28n-1%29 = 5%2Asqrt%282%29%5E%2818-1%29 = 5%2Asqrt%282%29%5E17 = 5*2^(17/2) = 5%2A2%5E8%2Asqrt%282%29 = 5%2A256%2Asqrt%282%29 = 1280%2Asqrt%282%29.



9. Find the common ratio of the geometric progression if its first term is  2  and the 11-th term is  486.

Solution.
Apply the formula  (1)  for the n-th term of a geometric progression.  You have

a%5B11%5D = a%5B1%5D%2Aq%5E%28n-1%29 = 2%2Aq%5E%2811-1%29 = 2%2Aq%5E10.

So, you get the equation

486 = 2%2Aq%5E10+

for the unknown common ratio q.  From this equation

q%5E10 = 486%2F2 = 243, or

q = root%2810%2C243%29 = root%2810%2C+3%5E5%29 = 3^(5/10) = 3^(1/2) = sqrt%283%29.

(note that 243 = 3%5E5 ).  Thus the common ratio is sqrt%283%29, and the progression is

2, 2sqrt%283%29, 6, 6sqrt%283%29, . . . .

Let us check that the 11-th term of the progression is 486.  Indeed, from the formula (1) for the n-th term of the progression you have

a%5B11%5D = a%5B1%5D%2Aq%5E%2811-1%29 = a%5B1%5D%2Aq%5E10 = 2%2A%28sqrt%283%29%29%5E10 = 2%2A3%5E5 = 2%2A243 = 486.


Formula for sum of a geometric progression

If  a, aq, aq%5E2, . . . , aq%5E%28n-1%29%5D  is the geometric progression with the first term  a  and the common ratio  q  then the formula for the sum of its first n terms is

      S%5Bn%5D = a+%2B+aq+%2B+aq%5E2+%2B+ellipsis+%2B+aq%5E%28n-1%29 = %28aq%5En+-+a%29%2F%28q-1%29.          (2)

Or, what is the same as
      S%5Bn%5D = a%2A%28%28q%5En+-+1%29%2F%28q-1%29%29.                                           (3)


For the proof of the formulas  (2),  (3)  see the lesson
    The proofs of the formulas for geometric progressions
under the current topic in this site.


Examples


8. Find the sum of the first  9  terms of the geometric progression  1, 2, 4, 8, . . . , 256.

Solution.
The geometric progression  1, 2, 3, 4, . . . , 256  has  1  as the first term,  2  as the common ratio and  9  terms in all.
Now, apply the formula (3) for the sum of the first  9  terms of an geometric progression.  You have

S%5B9%5D = 1+%2B+2+%2B+4+%2B+8+%2B+ellipsis+%2B+256 = 1%2A%28%282%5E9-1%29%2F%282-1%29%29 = 511%2F1 = 511.


9. Find the sum of the first  9  terms of the geometric progression  1, 1%2F2, 1%2F4, 1%2F8, . . . , 1%2F256.

Solution.
The geometric progression  1, 1%2F2, 1%2F4, 1%2F8, . . . , 1%2F256  has  1  as the first term,  1%2F2  as the common ratio and  9  terms in all.
Now, apply the formula  (3)  for the sum of the first  9 terms  of an geometric progression.  You have

S%5B9%5D = 1+%2B+1%2F2+%2B+1%2F4+%2B+1%2F8+%2B+ellipsis+%2B+1%2F256 = 1%2A%28%28%281%2F2%29%5E9-1%29%2F%281%2F2-1%29%29 = 2%2A%28%282%5E9-1%29%2F%282%5E9%29%29 = 511%2F256 = 2+-+1%2F256.


10. Find the sum of the first  9  terms of the geometric progression  1, -1%2F2, 1%2F4, -1%2F8, . . . , 1%2F256.

Solution.
The geometric progression  1,  -1%2F2,  1%2F4,  -1%2F8, . . . , 1%2F256  has  1  as the first term,  -1%2F2  as the common ratio and  9  terms in all.
Now, apply the formula  (3)  for the sum of the first  9  terms of an geometric progression.  You have

S%5B9%5D = 1+-+1%2F2+%2B+1%2F4+-+1%2F8+%2B+ellipsis+%2B+1%2F256 = 1%2A%28%28%28-1%2F2%29%5E9-1%29%2F%28-1%2F2-1%29%29 = %282%2A%282%5E9%2B1%29%2F%283%2A2%5E9%29%29 = %282%2F3%29%2A%28513%2F512%29 = %282%2F3%29%2A%281+%2B+1%2F512%29.


11. Find the sum of the first  10  terms of the geometric progression  1,  sqrt%282%29,  2,  2sqrt%282%29, . . . ,  %28sqrt%282%29%29%5E9+=+16sqrt%282%29.

Solution.
The geometric progression  1,  sqrt%282%29,  2,  2sqrt%282%29, . . . ,  %28sqrt%282%29%29%5E9+=+16sqrt%282%29  has  1  as the first term,  sqrt%282%29  as the common ratio and  10  terms in all.
Now, apply the formula  (3)  for the sum of the first  10  terms of an geometric progression. You have

S%5B10%5D = 1+%2B+sqrt%282%29+%2B+2+%2B+2sqrt%282%29+%2B+ellipsis+%2B+16sqrt%282%29 = 1%2A%28%28%28sqrt%282%29%29%5E10-1%29%2F%28sqrt%282%29-1%29%29 = 1%2A%28%282%5E5-1%29%2F%28sqrt%282%29-1%29%29 = 31%2F%28sqrt%282%29-1%29.


Summary


A  geometric progression  is a sequence of numbers such that the ratio of the current term to the preceding term is the same for any two consecutive terms.

The n-th term of the geometric progression with the first term  a  and the common ratio  q  is
    a%5Bn%5D = a%2Aq%5E%28n-1%29.

The sum of the first n terms of the geometric progression  a, aq, aq%5E2, ..., a%2Aq%5E%28n-1%29  with the first term  a  and the common ratio  q  is equal to
    S%5Bn%5D = %28aq%5En+-+a%29%2F%28q-1%29

or
    S%5Bn%5D = a%2A%28%28q%5En+-+1%29%2F%28q-1%29%29.


For some basic and typical problems on geometric progressions see the lesson  Problems on geometric progressions  under the current topic in this site.
For more word problems on geometric progressions see the lesson  Word problems on geometric progressions  under the current topic in this site.


My other lessons on geometric progressions in this site are
    - The proofs of the formulas for geometric progressions
    - Problems on geometric progressions
    - Solved problems on geometric progressions
    - Word problems on geometric progressions
    - One characteristic property of geometric progressions
    - Fresh, sweet and crispy problem on arithmetic and geometric progressions

    - Entertainment problems on geometric progressions

    - Ordinary Annuity saving plans and geometric progressions
    - Annuity Due saving plans and geometric progressions
    - Solved problems on Ordinary Annuity saving plans
    - Finding present value of an annuity, or an equivalent amount in today's dollars
    - Withdrawing a certain amount of money periodically from a compounded saving account
    - Miscellaneous problems on retirement plans

    - Mathematical induction and geometric progressions
    - Mathematical induction for sequences other than arithmetic or geometric

OVERVIEW of my lessons on geometric progressions with short annotations is in the lesson  OVERVIEW of lessons on geometric progressions.

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.

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