Find the common ratio and write out the first four terms of the geometric sequence bracket (3 to the power of n-1)/(10) bracket
???? please help me!
For n=1, 3^(n-1)/10 = 1/10
For n=2, 3^1/10 = 3/10
For n=3, 3^2/10 = 9/10
For n=4, 3^3/10 = 27/10
The ratio of successive terms is 3.
I don't know why you wrote "bracket" at the end of your question.
there are two brackets, one on each side of the problem. The ratio was right thank you! But it said the n=1-4 was wrong?
what is the answer for 15460 divide by ten?
To find the common ratio and write out the first four terms of the geometric sequence, we'll use the given formula:
aₙ = (3^(n-1))/10
Let's start with finding the common ratio (r) of the sequence.
In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio (r).
To find the common ratio, we can compare any two consecutive terms of the sequence. Let's compare the second term (a₂) with the first term (a₁):
a₂ = (3^(2-1))/10
a₂ = (3^1)/10
a₂ = 3/10
Since the second term is obtained by dividing the first term by 10, the common ratio is 1/10. Therefore, we have r = 1/10.
Now let's write out the first four terms of the sequence:
Term 1 (a₁):
a₁ = (3^(1-1))/10
a₁ = (3^0)/10
a₁ = 1/10
Term 2 (a₂):
a₂ = (3^(2-1))/10
a₂ = (3^1)/10
a₂ = 3/10
Term 3 (a₃):
a₃ = (3^(3-1))/10
a₃ = (3^2)/10
a₃ = 9/10
Term 4 (a₄):
a₄ = (3^(4-1))/10
a₄ = (3^3)/10
a₄ = 27/10
Therefore, the first four terms of the geometric sequence are:
1/10, 3/10, 9/10, 27/10