Lesson Problems on geometric progressions

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Problems on geometric progressions


This lesson shows you how to solve some typical problems on geometric progressions.

Problem 1

The third term of a geometric progression is  12  and the sixth term is  96.
Find the first term and the common ratio of the progression.

Solution

Let  a%5B1%5D  and  q  will be the first term and the common ratio of the progression respectively.
Use the formula for the  n-th term of a geometric progression

a%5Bn%5D = a%5B1%5Dq%5E%28n-1%29

(see the lesson  Geometric progressions  inder the current topic in this site).  You have

a%5B1%5Dq%5E2 = 12       for  n=3,  and
a%5B1%5Dq%5E5 = 96       for  n=6.

Now, divide the second equation by the first one (the left sides and the right sides). You get then

q%5E3= 96%2F12 = 8.

Hence,  q = 2.
To get the first term of the progression, substitute this value of  q  into the equation for the third term. You get

a%5B1%5D%2A2%5E2 = 12, or
a%5B1%5D = 12%2F4 = 3.

Thus the first term of the progression is  a%5B1%5D = 3 and the common ratio is  q = 2.
The first six terms of the progression are  3, 6, 12, 24, 48, 96.


Problem 2

The first term of a geometric progression is  7  and the fifth term is  567.  All the terms are positive.
Find the common ratio and the sum of the first  6  terms of the progression.

Solution

First, let us find the common ratio of the progression.
The formula for the fifth term of the geometric progression is

a%5B5%5D = a%5B1%5D%2Aq%5E%28n-1%29 = a%5B1%5D%2Aq%5E4.

(see the lesson  Geometric progressions  inder the current topic in this site).  Substitute the given data to this formula.  You get the equation for the unknown  q:

567 = 7%2Aq%5E4.

From this equation

q%5E4 = 567%2F7 = 81.

Hence,  q = root%284%2C81%29 = 3.  So, the common ratio is equal to  3.

Now, let us find the sum of the first  6  terms of this progression.  Use the formula for the sum of the first n terms of the geometric progression

S%5Bn%5D = a%5B1%5D+%2B+a%5B1%5Dq+%2B+a%5B1%5Dq%5E2+%2B+ellipsis+%2B+a%5B1%5Dq%5E%28n-1%29 = a%5B1%5D%2A%28q%5En+-+1%29%2F%28q-1%29.

(see the lesson  Geometric progressions  inder the current topic in this site).  Substitute  a%5B1%5D = 7,  q = 3,  n=6  to the formula.  You get

S%5B6%5D = 7%2A%283%5E6-1%29%2F%283-1%29 = 7%2A%286561-1%29%2F2 = 7%2A6560%2F2 = 7%2A3280 = 22960.

Answer. The common ratio of the geometric progression is  3, the sum of the first  6  terms is  22960.



Problem 3

The sum of the first and the third terms of the geometric progression is  20  and the sum of its first three terms is  26.
Find the progression.

Solution

Since the sum of the first and the third terms of the geometric progression is  20  and the sum of its first three terms is  26,  it is clear that the second term is  6.

Now, let  q  be the common difference of the progression. Then the first term is equal to  6%2Fq  and the third term is  6q.  Since the sum
of the first, the second and the third terms is  26, this leads to the equation

6%2Fq + 6 + 6q = 26

for the unknown  q.  Simplify this equation and transform it to the standard quadratic form step by step:

6+%2B+6q+%2B+6q%5E2+=+26q,
6q%5E2+-20q+%2B+6+=+0.

Apply the quadratic formula (see the lesson  Introduction into Quadratic Equations  in this site).  You have

q = %28-%28-20%29%2B-sqrt%2820%5E2+-+4%2A6%2A6%29%29%2F%282%2A6%29 = %2820%2B-sqrt%28400+-+144%29%29%2F12 = %2820%2B-sqrt%28256%29%29%2F12 = %2820%2B-16%29%2F12.

Hence, the roots of this quadratic equation are q%5B1%5D = %2820%2B16%29%2F12 = 36%2F12 = 3 and q%5B2%5D = %2820-16%29%2F12 = 4%2F12 = 1%2F3.

For the root  q%5B1%5D = 3  the first and the third terms of the progression are  6%2F3 = 2 and  6%2A3 = 18  respectively.
For the root  q%5B1%5D = 1%2F3  the first and the third terms of the progression are  6%2F%281%2F3%29 = 6%2A3 = 18  and 6%2A%281%2F3%29 = 2.
In both cases, we have the same set of numbers  2, 6 and 18.

Answer. The progression is  2, 6, 18  or  18, 6, 2.


Problem 4

Let  a%5B1%5D, a%5B2%5D, a%5B3%5D, . . . , a%5Bn%5D  be a geometric progression with the sum equal to S.
Find the sum of the sequence  1%2Fa%5B1%5D,  1%2Fa%5B2%5D,  1%2Fa%5B3%5D, . . . ,  1%2Fa%5Bn%5D.

Solution

Let  q  be the common ratio of the geometric progression  a%5B1%5D, a%5B2%5D,, a%5B3%5D, . . . , a%5Bn%5D.
Note that the sequence  1%2Fa%5B1%5D, 1%2Fa%5B2%5D, 1%2Fa%5B3%5D, . . . , 1%2Fa%5Bn%5D  is the geometric progression with the common ratio 1%2Fq.
Indeed, for any two consecutive terms of this progression  1%2Fa%5Bk-1%5D,  1%2Fa%5Bk%5D  the ratios of the current term  1%2Fa%5Bk%5D  to the preceding one  1%2Fa%5Bk-1%5D  is equal to
1%2Fa%5Bk%5D : 1%2Fa%5Bk-1%5D = a%5Bk-1%5D%2Fa%5Bk%5D = 1%2Fq,
that is the constant value.  Thus the sequence  1%2Fa%5B1%5D, 1%2Fa%5B2%5D, 1%2Fa%5B3%5D, . . . , 1%2Fa%5Bn%5D  is the geometric progression with the first term  1%2Fa%5B1%5D  and the common ratio  1%2Fq.
According to the formula for the sum of the first n terms of a geometric progression applied to the geometric progression  1%2Fa%5B1%5D, 1%2Fa%5B2%5D, 1%2Fa%5B3%5D, . . . , 1%2Fa%5Bn%5D,  its sum  T  is equal to
T = 1%2Fa%5B1%5D*%28%281%2Fq%29%5En+-+1%29%2F%281%2Fq-1%29 = 1%2Fa%5B1%5D*%28q%5En-1%29%2F%28q%5En%2A%28q-1%29%29%2Aq.

Now, you can group the terms as follows:

T = 1%2Fa%5B1%5D*%28q%5En-1%29%2F%28q%5En%2A%28q-1%29%29%2Aq = 1%2F%28a%5B1%5D%2Aq%5E%28n-1%29%29*%28q%5En-1%29%2F%28q-1%29 = 1%2Fa%5B1%5D*1%2F%28a%5B1%5D%2Aq%5E%28n-1%29%29*%28a%5B1%5D%2A%28q%5En-1%29%2F%28q-1%29%29.
                                                               ~~~~~~~~~  ~~~~~~~~~~
Note that the first underlined multiplier in the right side is nothing else as  1%2Fa%5Bn%5D,  while the second underlined multiplier is the sum  S  of the original geometric
progression  a%5B1%5D, a%5B2%5D, a%5B3%5D, . . . , a%5Bn%5D.  Thus the answer is

T = S%2F%28a%5B1%5D%2Aa%5Bn%5D%29.


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
    - Geometric progressions
    - The proofs of the formulas for 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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