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Let the m-th and n-th terms of a geometric progression be 3/4 and 12 , respectively, where m<n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n - m is
- a)-2
- b)2
- c)6
- d)4
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let the m-th and n-th terms of a geometric progression be 3/4and 12 , ...
To get the minimum value for r + n -m, r should be minimum.
∴ r = - 4
n -m = 2
∴ Required answer =-2
∴ r = - 4
n -m = 2
∴ Required answer =-2
Most Upvoted Answer
Let the m-th and n-th terms of a geometric progression be 3/4and 12 , ...
Let the first term of the geometric progression be a and the common ratio be r.
The m-th term can be written as: a * r^(m-1) = 3/4
The n-th term can be written as: a * r^(n-1) = 12
Dividing the two equations, we get:
(a * r^(m-1))/(a * r^(n-1)) = (3/4)/12
r^(m-n) = (3/4)/12
r^(m-n) = 1/48
Taking the logarithm of both sides, we get:
(m-n) * log(r) = log(1/48)
Since we know that m > n, we can solve for (m-n):
(m-n) = log(1/48)/log(r)
So, (m-n) is equal to log(1/48)/log(r).
Therefore, the difference between m and n is log(1/48)/log(r).
The m-th term can be written as: a * r^(m-1) = 3/4
The n-th term can be written as: a * r^(n-1) = 12
Dividing the two equations, we get:
(a * r^(m-1))/(a * r^(n-1)) = (3/4)/12
r^(m-n) = (3/4)/12
r^(m-n) = 1/48
Taking the logarithm of both sides, we get:
(m-n) * log(r) = log(1/48)
Since we know that m > n, we can solve for (m-n):
(m-n) = log(1/48)/log(r)
So, (m-n) is equal to log(1/48)/log(r).
Therefore, the difference between m and n is log(1/48)/log(r).
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