# How to Find a Certain Term, Given Two Terms in a Geometric Sequence?

The nth Term Of A Geometric Sequence
If {an} is a geometric sequence with common ratio r, then

a2=a1r
a3=a2r=a1r2
a4=a3r=a1r3

an=a1r(n-1) for every n>1

Example 1: Finding a Term in a Geometric Sequence
In a geometric sequence, the third term is 24 and the 6th term is 192. Find the 10th term. Solution: Here, a3=a1r2=24 … (1)
and a6=a1r5=192 … (2)
Dividing (2) by (1), we get r=2. Substituting r=2 in (1), we get a=6.
Hence a10=6⋅29=3072.

Example 2. If the first and tenth terms of a geometric sequence are 1 and 4, find the seventeenth term to three decimal places.
Solution:
First let n=10, a1=1, a10=4 and use the formula an=a1r(n-1) to find r.

an=a1r(n-1)
a10=1⋅r(10-1)
4=r9r=4(1⁄9)

Now use the formula an=a1r(n-1) again, this time with n=17.

a17=a1r17=1(4(1⁄9))17=4(17⁄9)≈13.716

If we know the first term in a geometric progression and the ratio between successive terms, then we can work out the value of any term in the geometric progression . The nth term is given by

an=a1r(n-1)

Again, a is the first term and r is the ratio. Remember that a1r(n-1)≠(a1r)(n-1).

Question 1. Given the first two terms in a geometric progression as 2 and 4, what is the 10th term?
Solution:

a1=2;r=4/2=2

Then a10=2×29=1024.

Question 2. Given the first two terms in a geometric progression as 5 and ½, what is the 7th term?
Solution: Example 3. Find the 20th and nth terms of the geometric sequence 5/2, 5/4, ⅝, …
Solution:
The given geometric sequence is 5/2, 5/4, ⅝, … .
Here, a1= First term =5/2. Question 3. (Challenge) The fifth term of a geometric sequence is 1/27 of the eighth term. If the ninth term is 702, what is the eighth term?
Solution:
Given a5=a8/27 and a9=702.
Find the Value of r. Find the Value of a8.
a9=a8r
702=a8⋅3
a8=234

Example 4: Finding Terms in a Geometric Sequence
If the third term of a geometric sequence is -12 and the fourth term is 24, find the first and fifth terms of the sequence.
Solution:
Divide the 4th term by the 3rd term to find the common ratio.
The common ratio is 24/(-12) or -2. Substitute 3 for n and -2 for r to find the first term.

an=a1r(n-1)
a3=a1⋅(-2)(3-1)
-12=a1⋅(-2)2
-12=4a1
-3=a1

The first term is -3. Find the fifth term.
an=a1r(n-1)
a5=-3(-2)(5-1)
=-3(-2)4
a5=-48

The fifth term is -48.

Example 5: Determining Terms and the Number of Terms in a Finite Geometric Sequence
In a finite geometric sequence, t1=5 and t5=1280
a) Determine t2 and t6.
b) The last term of the sequence is 20, 480. How many terms are in the sequence?
solution:
a) Determine the common ratio.
Use: tn=t1 r(n-1)
Substitute: n=5, t5=1280, t1=5

1280=5r(5-1)

Simplify.
1280=5r4

Divide each side by 5.
256=r4

Take the fourth root of each side.
16=r2
r=-4 or r=4

There are 2 possible values for r.
(i) When r=-4, then t2 is 5(4)=20
(ii) When r=4, then t2 is 5(-4)=-20
To determine t6, use: tn=t1 r(n-1)
(i) Substitute: n=6, t1=5, r=-4
t6=5(-4)(6-1)=5(-4)5=-5120

(ii) Substitute: n=6, t1=5, r=4
t6=5⋅4(6-1)=5⋅45=5120

So, t2 is -20 or 20, and t6 is -5120 or 5120.

b) Since the last term is positive, use the positive value of r.

tn=t1 r(n-1)

Substitute: tn=20 480, t1=5, r=4
20480=5⋅4(n-1)

Divide each side by 5.
4096=4(n-1)

Use guess and test to determine which power of 4 is equal to 4096.
Guess: 44=256
This is too low. Guess: 46=4096
This is correct.
So, 46=4(n-1)
Equate exponents.
6=n-1
n=7

There are 7 terms in the sequence.

Ratio of two terms of Geometric Sequence
Let the geometric sequence be

a, ar, a1r2, …

Now, Example 6. The 5th, 8th and 11th terms of a geometric sequence are p, q and s, respectively. Show that q2=p⋅s.
Solution:
Let a be the first term and r be the common ratio of the geometric sequence
According to the given condition,

a5=a1r(5-1)=a1r4=p … (1)
a8=a1r(8-1)=a1r7=q … (2)
a11=a1r(11-1)=ar10=s … (3)

Dividing equation (2) by (1), we obtain Dividing equation (3) by (2), we obtain Equating the values of r3 obtained in (4) and (5), we obtain Thus, the given result has proven.

RELATED POSTs