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If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ?
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If the first and the nth term of a G.P are a and b , respectively , an...
Proof:

Given: First term of G.P = a and nth term of G.P = b
To prove: P*2 = [ab]*n

Step 1: Find the common ratio of the G.P
Let the common ratio of the G.P be r.
Then, we know that b = ar^(n-1)

Step 2: Find the product of the first and nth terms of the G.P
The product of the first and nth terms of the G.P is ab.

Step 3: Find the product of the G.P
The product of the G.P can be written as P = a * ar * ar^2 * ... * ar^(n-1) = a^n * r^(1+2+...+(n-1)) = a^n * r^(n*(n-1)/2)

Step 4: Simplify the expression for the product of the G.P
We know that r^(n-1) = b/a, so r^n = b/a * r.
Substituting this in the expression for P, we get:
P = a^n * r^(n*(n-1)/2) = a^n * (b/a)^((n-1)/2) * r^(n*(n-1)/2)
= b^((n-1)/2) * (a*r^(n-1))^((n-1)/2)
= b^((n-1)/2) * b^((n-1)/2)
= b^n

Step 5: Prove P*2 = [ab]*n
P*2 = 2b^n (from Step 4)
[ab]*n = ab * n * r^(n-1) = ab * n * (b/a) = b^n * n
Since P*2 = b^n * 2, we can see that P*2 = [ab]*n.

Hence, we have proved that P*2 = [ab]*n.
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If the first and the nth term of a G.P are a and b , respectively , and if P is the product of n terms , prove that P*2 = [ab]*n ?
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