A geometric sequence has a constant ratio (common ratio) between consecutive terms.
For 3, 9, 27, ... the common ratio is 3 because:
3 X 3 = 9
9 X 3 = 27
So to find the 7th term you can do it two ways:
One way:
3 is the 1st term, 9 is the 2nd term, 27 is the 3rd term so then
4th term: 27 X 3 = 81
5th term: 81 X 3 = 243
6th term: 243 X 3 = 729
7th term: 729 X 3 = 2,187
Another way:
You can use the explicit formula #a_n = a_1 *r^(n-1)#, where #a_n# is the nth term, #a_1# is the first term, n is the number of the term, and r is the common ratio
so #a_7 = 3 * 3^ (7-1)#
#a_7 = 3 * 3^ (6)#
#a_7 = 3 * 729#
#a_7 = 2,187#
Both ways get you to the same answer that the 7th term in that geometric sequence is 2, 187 .
Solution:
The nth term of the geometric sequence is given by, an = a rn - 1,
Where a and r being the first term and the common ratio respectively.
Given a1 = a = 128 and a3 = 8
We know an = a rn - 1,
⇒ a3 = a.r3 - 1
⇒ 8 = 128 × r2
⇒ r2 = 8/128
⇒ r2 = 1/16
⇒ r = ±1/4
Now, a7 = a.r7 - 1 = a.r6
⇒ a7 = 128 × (1/4)6
⇒ a7 = 128/4096
⇒ a7 = 1/32
If r = -1/4,
⇒ a7 = 128 × (-1/4)6
⇒ a7 = 128/4096
⇒ a7 = 1/32
Therefore, the 7th term of the geometric sequence a7 is 1/32.
What is the 7th term of the geometric sequence where a1 = 128 and a3 = 8?
Summary:
The 7th term of the geometric sequence where a1 = 128 and a3 = 8 is 1/32.
Algebra Examples
Popular Problems
Algebra
Find the 7th Term 1 , 2 , 4 , 8 , 16 , 32
, , , , ,
Step 1
This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by gives the next term. In other words, .
Geometric Sequence:
Step 2
This is the form of a geometric sequence.
Step 3
Substitute in the values of and .
Step 4
Multiply by .
Step 5
Substitute in the value of to find the th term.
Step 6
Subtract from .
Step 7
Raise to the power of .
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Solved.
On geometric progressions, see the lessons
- Geometric progressions
-
The proofs of the formulas for geometric progressions
- Problems on geometric progressions
- Word problems on geometric progressions
- One characteristic property of geometric progressions
-
Solved problems on geometric progressions
- Fresh, sweet and crispy problem on arithmetic and geometric progressions
Also,
you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Geometric progressions".