**General Equation for the nth-term of a Geometric Sequence**

From the above example we know

*a*

_{1}=2 and

*r*=2, and we have seen from the table that the

*n*th-term is given by

*a*=2×2

_{n}^{(n-1)}. Thus, in general,

*a*=

_{n}*a*

_{1}⋅

*r*

^{(n-1)}

where

*a*

_{1}is the first term and

*r*is called the

__common ratio__.

So, if we want to know how many people are newly-infected after 10 days, we need to work out *a*_{10}.

*a*=

_{n}*a*

_{1}⋅

*r*

^{(n-1)}

*a*

_{10}=2⋅

*r*

^{(10-1)}

2×2

^{9}=2×512=1024.

That is, after 10 days, there are 1024 newly-infected people.

Or, how many days would pass before 16, 384 people become newly infected with the flu virus?

*a*=

_{n}*a*

_{1}⋅

*r*

^{(n-1)}

16, 384=2⋅2

^{(n-1)}

16, 384÷2=2

^{(n-1)}

8 192=2

^{(n-1)}

2

^{13}=2

^{(n-1)}

13=

*n*-1

*n*=14.

That is, 14 days pass before 16, 384 people are newly-infected.

📌 Example 1. Determine the number of terms in the sequence: 2, 4, 8, …, 1024.

✍ Solution:

*r*=4/2=8/4=2

*a*⋅

*r*

^{(n-1)}=1024

2⋅2

^{(n-1)}=2

^{10}→2

^{n}=2

^{10}

*n*=10

📌 Ex2. Find *n* in the geometric sequence 4, -2, 1, …. if *a _{n}*=1/16

✍ Solution:

Since 4, -2, 1, …

Here,

*a*=4,

*r*=-2÷4=-½,

*a*=1/16

_{n}*a*=

_{n}*a*⋅

*r*

^{(n-1)}

1/16=4⋅(-½)

^{(n-1)}

1/64=(-½)

^{(n-1)}

(-½)

^{6}=(-½)

^{(n-1)}

6=

*n*-1→

*n*=7.

We can also check whether a given number belongs to a given geometric sequence.

📌 Question 1. Does the number 48 belong to the geometric sequence

✍ Solution:

Here

*a*=3072 and

*r*=½, so

*a*=3072×½

_{n}^{(n-1)}.

We set 3072×½

^{(n-1)}=48. This gives ½

^{(n-1)}=1/64, that is, 2

^{(n-1)}=64=2

^{6}, and so

*n*=7.

Hence, 48 is the 7th term of the sequence.

📌 Q2. Does the number 6072 belong to the geometric sequence

✍ Solution:

Here

*a*=3 and

*r*=-2, so

*a*=3×(-2)

_{n}^{(n-1)}.

We set 3×(-2)

^{(n-1)}=6072. This gives (-2)

^{(n-1)}=2024.

But 2024 is not a power of 2, and so 6072 does not belong to the sequence.

If given a sequence then check whether it is either arithmetic, geometric or quadratic.

📌 Q3. Given the sequence 1, ⅔, 9/4, ….

a) Determine the next two terms

b) Which term of the sequence is equal to 32/243?

✍ Solution:

The common ratio ⅔ because ⅔÷1=⅔=4/9÷⅔

a) *a*_{4}=*a*⋅r^{3}=1⋅⅔^{3}=8/27 and *a*_{5}=*a*⋅r^{4}=1⋅⅔^{4}=16/81

b) *a*=1, *r*=⅔ and *a _{n}*=

*a*⋅

*r*

^{(n-1)}=32/243

^{5}

*n*-1=5

*n*=6.

📌 Q4: Which term of the geometric sequence., 2, 8, 32, up to *n* terms is 131072?

✍ Solution: Let 131072 be the *n*th term of the given geometric sequence. Here *a*=2 and *r*=4. Therefore 131072=*a _{n}*=2⋅4

^{(n-1)}or 65536=4

^{(n-1)}

This gives 4

^{8}=4

^{(n-1)}

So that

*n*-1=8, ie,

*n*=9. Hence, 131072 is the 9th term of the geometric sequence.

📌 Questions:

(a) What term number is 128 of the sequence 2, 2√2, 4, …?

(b) What term number is 729 of the sequence √3, 3, 3√3, …?

(c) What term number is 1/19683 of the sequence ⅓, 1/9, 1/27, …?

✍ Solutions:

(a) *a*_{1}=(√2)^{2}; *a*_{2}=(√2)^{3}; *a*_{3}=(√2)^{4}; *a _{n}*=(√2)

^{(n+1)}

^{7}=(√2)

^{(2×7)}=(√2)

^{(13+1)}=

*a*

_{13}

Thus, 128 is the 13th term of the given sequence.

(b)

*a*

_{1}=(√3)

^{1};

*a*

_{2}=(√3)

^{2};

*a*

_{3}=(√3)

^{3};

*a*=(√3)

_{n}^{n}

^{6}=(√3)

^{(2×6)}=(√2)

^{12}=

*a*

_{12}

Thus, 729 is the 12th term of the given sequence.

(c)

*a*

_{1}=⅓

^{1};

*a*

_{2}=⅓

^{2};

*a*

_{3}=⅓

^{3};

*a*=⅓

_{n}^{n}

1/19683=⅓

^{9}=

*a*

_{9}.

Thus, 1/19683 is the 9th term of the given sequence.