Geometric Sequences Problems with Solutions

Problems and exercises involving geometric sequences, along with detailed solutions and answers, are presented.

REVIEW OF GEOMETRIC SEQUENCES

The sequence shown below

2 , 8 , 32 , 128 , ...


has been obtained starting from 2 and multiplying each term by 4. 2 is the first term of the sequence and 4 is the common ratio.

8 = 2 * 4 or 8 / 2 = 4

32 = 8 * 4 or 32 / 8 = 4

128 = 32 * 4 or 128 / 32 = 4 and so on.

The terms in the sequence may also be written as follows

a
1 = 2

a
2 = a1 * 4 = 2 * 4

a
3 = a2 * 4 = 2 * 42

a
4 = a3 * 4 = 2 * 43

The n th term may now be written as

an = a1 * rn-1


where a
1 is the first term of the sequence and r is the common ratio which is equal to 4 in the above example.

The sum of the first n terms of a geometric sequence is given by

sn = a1 + a2 + a3 + ... an = a1 (1 - rn) / (1 - r)


The sum S of an infinite (n approaches infinity) geometric sequence and when |r| <1 is given by

S = a1 / (1 - r)


Problem 1:
Find the terms a2, a3, a4 and a5 of a geometric sequence if a1 = 10 and the common ratio r = - 1.

Solution to Problem 1:

  • Use the definition of a geometric sequence

    a2 = a1 * r = 10 * (-1) = - 10

    a3 = a2 * r = - 10 * (-1) = 10

    a4 = a3 * r = 10 * (-1) = - 10

    a5 = a4 * r = - 10 * (-1) = 10


Problem 2:
Find the 10 th term of a geometric sequence if a1 = 45 and the common ration r = 0.2.

Solution to Problem 2:

  • Use the formula an = a1 * rn-1 that gives the n th term to find a10 as follows

    a10 = a1 * rn-1

    = 45 * 0.29 = 2.304 * 10-5


Problem 3:
Find a20 of a geometric sequence if the first few terms of the sequence are given by

-1/2 , 1/4 , -1/8 , 1 / 16 , ...

Solution to Problem 3:

  • We first use the first few terms to find the common ratio

    r = a2 / a1 = (1/4) / (-1/2) = -1/2

    r = a3 / a2 = (-1/8) / (1/4) = -1/2

    r = a4 / a3 = (1/16) / (-1/8) = -1/2

  • The common ration r = -1/2. We now use the formula an = a1 * rn-1 for the n th term to find a20 as follows.

    a20 = a1 * r20-1

    = (-1/2) * (-1/2)20-1 = 1 / (2020)


Problem 4:

Given the terms a10 = 3 / 512 and a15 = 3 / 16384 of a geometric sequence, find the exact value of the term a30 of the sequence.

Solution to Problem 4:

  • We first use the formula for the n th term to write a10 and a15 as follows

    a10 = a1 * r10-1 = 3 / 512

    a15 = a1 * r15-1 = 3 / 16384

  • We now divide the terms a10 and a15 to write

    a15 / a10 = (a1 * r14 / a1 * r9) = (3 / 16384) / (3 / 512)

  • Solve for r to obtain.

    r5 = 1 / 32 which gives r = 1/2

  • We now use a10 to find a1 as follows.

    a10 = 3 / 512 = a1(1/2)9

  • Solve for a1 to obtain.

    a1 = 3

  • We now use the formula for the n th term to find a30 as follows.

    a30 = 3(1/2)29 = 3 / 536870912


Problem 5:
Find the sum

6
S = 3k-1
k=1

Solution to Problem 5:

  • We first rewrite the sum S as follows

    S = 1 + 3 + 9 + 27 + 81 + 243 = 364

  • Another method is to first note that the terms making the sum are those of an arithmetic sequence with a1 = 1 and r = 3 using the formula sn = a1 (1 - rn) / (1 - r) with n = 6.

    s6 = 1 (1 - 36) / (1 - 3) = 364


Problem 6:
Find the sum

10
S = 8*(1/4)i-1
i=1

Solution to Problem 6:

  • An examination of the terms included in the sum are

    8 , 8*((1/4)1 , 8*((1/4)2 , ... , 8*((1/4)9

  • These are the terms of a geometric sequence with a1 = 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence

    s10 = a1 (1 - rn) / (1 - r)

    = 8 * (1 - (1/4)10) / (1 - 1/4) = 10.67 (rounded to 2 decimal places)


Problem 7:
Write the rational number 5.31313131... as the ratio of two integers.

Solution to Problem 7:

  • We first write the given rational number as an infinite sum as follows

    5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + ....

  • The terms making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a1 = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence

    S = a1 / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99

  • We now write 5.313131... as follows

    5.313131... = 5 + 31/99 = 526 / 99



Exercises:
Answer the following questions related to geometric sequences:

a) Find a
20 given that a3 = 1/2 and a5 = 8

b) Find a
30 given that the first few terms of a geometric sequence are given by -2 , 1 , -1/2 , 1/4 ...

c) Find r given that a
1 = 10 and a20 = 10-18

d) write the rational number 0.9717171... as a ratio of two positive integers.

Solutions to Above Exercises:

a) a
20 = 218

b) a
30 = 1 / 228

c) r = 0.1

d) 0.9717171... = 481/495

More math problems with detailed solutions in this site.

Arithmetic Series Online Calculator. An online calculator to calculate the sum of the terms in an arithmetic sequence.



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Updated: 2 April 2013

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