The 4th and 6th terms of a G.P are 1/10 and 1/160 respectively determine the common ratio and the first term
Let the first term of the G.P be 'a' and the common ratio be 'r'.
We know that the 4th term of the G.P is 1/10, so:
a * r^3 = 1/10
We also know that the 6th term of the G.P is 1/160, so:
a * r^5 = 1/160
Dividing the second equation by the first equation, we get:
r^2 = (1/160) / (1/10)
r^2 = 1/16
r = 1/4 (since r is positive)
Substituting this value of r in the first equation, we get:
a * (1/4)^3 = 1/10
a * 1/64 = 1/10
a = (1/10) * 64
a = 6.4
Therefore, the common ratio is 1/4 and the first term is 6.4.
Why did the geometric progression go to therapy? Because it felt like it was stuck in an infinite loop of numbers!
Let's solve this riddle. We'll use the formula for the nth term of a geometric progression:
tn = a * (r^(n-1))
Given that the 4th term (t4) is 1/10 and the 6th term (t6) is 1/160, we can set up two equations using this formula:
1/10 = a * (r^(4-1))
1/160 = a * (r^(6-1))
Simplifying both equations, we have:
1/10 = a * r^3
1/160 = a * r^5
Now, to find the common ratio (r), we can divide the two equations:
(1/10) / (1/160) = (a * r^3) / (a * r^5)
Simplifying further:
16 = r^2
Taking the square root of both sides, we get:
r = ±√16
Since negative values for r wouldn't make sense in this context, we are left with:
r = √16 = 4
Now, let's find the first term (a) by substituting the value of r into one of the equations:
1/10 = a * (4^(4-1))
1/10 = a * 4^3
1/10 = a * 64
a = (1/10) * (1/64)
a = 1/640
So, the common ratio (r) is 4 and the first term (a) is 1/640.
To find the common ratio and the first term of a geometric progression (G.P.), we can use the formula for the nth term of a G.P., which is given as:
an = a * r^(n-1)
where:
an denotes the nth term of the G.P.
a represents the first term of the G.P.
r represents the common ratio of the G.P.
n represents the position of the term in the G.P.
Given that the 4th term (a4) is 1/10 and the 6th term (a6) is 1/160, we can substitute these values into the formula to form two equations:
a4 = a * r^(4-1) ...(1)
a6 = a * r^(6-1) ...(2)
Substituting the values of the 4th and 6th terms:
1/10 = a * r^3 ...(1)
1/160 = a * r^5 ...(2)
To find the common ratio (r), we divide equation (2) by equation (1):
(1/160) / (1/10) = (a * r^5) / (a * r^3)
1/16 = r^2
Taking the square root of both sides:
r = ±√(1/16)
r = ±1/4
Since the common ratio cannot be negative in a G.P., we can conclude that the common ratio (r) is 1/4.
To find the first term (a), we can substitute the value of r into equation (1):
1/10 = a * (1/4)^3
1/10 = a * 1/64
64/10 = a
a = 32/5
Therefore, the common ratio is 1/4 and the first term is 32/5.
To find the common ratio and the first term of a geometric progression (G.P.), we can use the formula for the nth term of a G.P.:
aₙ = a₁ * r^(n-1)
where aₙ is the nth term, a₁ is the first term, r is the common ratio, and n is the term number.
Given that the 4th term (a₄) is 1/10 and the 6th term (a₆) is 1/160, we can set up the equations as follows:
a₄ = a₁ * r^(4-1) = 1/10
a₆ = a₁ * r^(6-1) = 1/160
We have two equations with two unknowns (a₁ and r). Let's solve them step by step:
1) Start with the equation a₄ = a₁ * r^(4-1) = 1/10:
a₁ * r^3 = 1/10 ---(equation 1)
2) Similarly, for a₆ = a₁ * r^(6-1) = 1/160:
a₁ * r^5 = 1/160 ---(equation 2)
3) Dividing equation 2 by equation 1, we can cancel out the term "a₁":
(r^5)/(r^3) = (1/160)/(1/10)
r^(5-3) = (1/160)*(10/1)
r^2 = 1/16
r = √(1/16)
r = 1/4 ---(equation 3)
4) Substituting the value of r (1/4) into equation 1 to solve for a₁:
a₁ * (1/4)^3 = 1/10
a₁ * 1/64 = 1/10
a₁ = (1/10) * 64
a₁ = 64/10
a₁ = 32/5 ---(equation 4)
Hence, the common ratio (r) is 1/4 and the first term (a₁) is 32/5.