Write the geometric sequence that has five geometric means between 1and 15,625.
I got:
1, 5, 25, 125, 625, 3,125, 15,625, …
is it ok?
you are correct if your last numbers are 3125 and 15625.
find the perpendicular distance from the corner of the triangle to the opposite side if all the sides 25cm long
calculate
Well, your sequence seems to be missing a few terms! Let me help you out with a proper geometric sequence:
1, 5, 25, 125, 625, 3125, 15,625.
Now that's a complete sequence! Don't you love it when things come together? Keep being geometrically awesome!
Your sequence seems to be incorrect. Let me explain how to find the correct geometric sequence with five geometric means between 1 and 15,625.
To find a geometric sequence with a given first term, last term, and number of geometric means, we need to use the formula:
an = a * r^(n-1)
where:
an = the nth term,
a = the first term,
r = the common ratio, and
n = the term number.
In this case, we have the first term (a = 1) and the last term (an = 15,625). We also have five geometric means, which means we have five terms between the first and last term.
Since we have one more term after the last geometric mean, the total number of terms would be six. Therefore, n = 6.
To find the common ratio (r), we can use the formula:
r = (an / a)^(1/(n-1))
Substituting the values, we get:
r = (15,625 / 1)^(1/(6-1))
r = 15,625^(1/5)
r = 5
Now, we can use the formula to find the terms of the sequence:
a1 = 1 * 5^(1-1) = 1
a2 = 1 * 5^(2-1) = 5
a3 = 1 * 5^(3-1) = 25
a4 = 1 * 5^(4-1) = 125
a5 = 1 * 5^(5-1) = 625
a6 = 1 * 5^(6-1) = 3,125
Therefore, the correct geometric sequence with five geometric means between 1 and 15,625 is:
1, 5, 25, 125, 625, 3,125
Partially correct.
Let's the first term in the sequence the 1,
So
a1 = 1
If the ratio between terms is r, then any term in the sequence is :
an = a1 ⋅ r ⁿ⁻¹
If there are 5 geometric means between 1 and 15 625, then 15 625 must be the 7th term ( a7 ).
a7 = 1 ⋅ r⁷⁻¹
15 625 = r⁶
r = ⁶√ 15 625
r = ± 5
a1 = 1
r = 5
a1 = 1 ∙ 5⁰ = 1 ∙ 1 = 1
a2 = 1 ∙ 5¹ = 1 ∙ 5 = 5
a3 = 1 ∙ 5² = 1 ∙ 25 = 25
a4 = 1 ∙ 5³ = 1 ∙ 125 = 125
a5 = 1 ∙ 5⁴ = 1 ∙ 625 = 625
a6 = 1 ∙ 5⁵ = 1 ∙ 3 125 = 3 125
a7 = 1 ∙ 5⁶ = 1 ∙ 15 625 = 15 625
1 , 5 , 25 , 125 , 625 , 3 125 , 15 625
and
a1 = 1
r = - 5
a1 = 1 ∙ ( - 5 )⁰ = 1 ∙ 1 = 1
a2 = 1 ∙ ( - 5 )¹ = 1 ∙ ( - 5 ) = - 5
a3 = 1 ∙ ( - 5 )² = 1 ∙ 25 = 25
a4 = 1 ∙ ( - 5 )³ = 1 ∙ ( -125 ) = - 125
a5 = 1 ∙ ( - 5 )⁴ = 1 ∙ 625 = 625
a6 = 1 ∙ ( - 5 )⁵ = 1 ∙ ( - 3 125 ) = - 3 125
a7 = 1 ∙ ( - 5 )⁶ = 1 ∙ 15 625 = 15 625
1 , - 5 , 25 , - 125 , 625 , - 3125 , 15 625