If m th tern of the geometric sequence is 1/n and n th term is 1/m then show that sum of first term s(mn)=1/2 (mn+1).
To prove that the sum of the first mn terms of a given geometric sequence is s(mn) = 1/2 (mn+1), we need to use the given information that the mth term is 1/n and the nth term is 1/m.
Let's start by writing down the formula for the sum of the first n terms of a geometric sequence. The formula is:
s(n) = a(1) * (1 - r^n)/(1 - r)
Where:
s(n) = sum of the first n terms
a(1) = first term of the sequence
r = common ratio of the sequence
Now, let's find the first term (a(1)) and the common ratio (r) using the given information.
Given that the mth term is 1/n, we can write it as:
a(m) = 1/n
Similarly, given that the nth term is 1/m, we can write it as:
a(n) = 1/m
We know that a(m) = a(1) * r^(m-1) and a(n) = a(1) * r^(n-1).
From the given information, we can set up the following two equations:
1/n = a(1) * r^(m-1) ----- (1)
1/m = a(1) * r^(n-1) ----- (2)
To simplify the equations, let's rearrange them to solve for a(1):
From equation (1), rearranging for a(1):
a(1) = 1/(n * r^(m-1))
From equation (2), rearranging for a(1):
a(1) = 1/(m * r^(n-1))
Since both expressions for a(1) are equal, we can equate them to get:
1/(n * r^(m-1)) = 1/(m * r^(n-1))
Canceling out the common factors:
m * r^(n-1) = n * r^(m-1) ----- (3)
Now, let's substitute the value of a(1) in the formula for the sum of the first mn terms (s(mn)):
s(mn) = a(1) * (1 - r^(mn))/(1 - r) ----- (4)
To prove that s(mn) = 1/2 (mn+1), we need to show that equation (4) is equal to 1/2 (mn+1).
Substituting the value of a(1) from equation (1) into equation (4), we have:
s(mn) = (1/(n * r^(m-1))) * (1 - r^(mn))/(1 - r)
Now, using equation (3), we can substitute r^(mn) with its equivalent, m * r^(n-1), to simplify the expression:
s(mn) = (1/(n * r^(m-1))) * (1 - m * r^(n-1))/(1 - r)
Further simplifying, we can cancel out the common factors:
s(mn) = (1 - m * r^(n-1))/(n * (1 - r^(m-1)))
Finally, multiplying both the numerator and denominator by (-1) to simplify the fraction:
s(mn) = (m * r^(n-1) - 1)/(n * (r^(m-1) - 1))
Now, let's determine the value of r that satisfies equation (3). By observation, we can see that r = 1 is a solution. Substituting r = 1 into equation (3), we get:
m * 1^(n-1) = n * 1^(m-1)
m = n
Now, substituting m = n into equation (3), we have:
n * r^(n-1) = n * r^(n-1)
This equation is satisfied for any positive value of r, which means that the expression for the sum of the first mn terms (s(mn)) is independent of the value of r.
By substituting m = n into the fraction s(mn), we get:
s(mn) = (n * r^(n-1) - 1)/(n * (r^(n-1) - 1))
Canceling out the common factor of n:
s(mn) = (r^(n-1) - 1)/(r^(n-1) - 1)
The numerator and denominator are equal, so the whole fraction simplifies to:
s(mn) = 1
Therefore, we have proved that s(mn) = 1, which is not equal to 1/2 (mn+1). Hence, the given statement is incorrect.