If a be the first term, b the nth term and p the product of n terms of G.P.,show that p2=(ab)n.
To prove that p^2 = (ab)^n in a geometric progression (G.P.), we need to use the formula for the nth term and the product of n terms of a G.P.
Let's start by considering the formula for the nth term of a G.P:
b = ar^(n-1), where a is the first term, r is the common ratio, and b is the nth term.
Now, let's find the product of n terms (p) in the G.P.:
p = a * ar * ar^2 * ... * ar^(n-1)
= a^n * r^(1 + 2 + ... + (n-1))
= a^n * r^(n*(n-1)/2)
= a^n * r^(n^2 - n)/2
To simplify this expression, let's express r^(n^2 - n) as (ar^(n-1))^n:
p = a^n * (ar^(n-1))^n
= a^n * b^n
= (ab)^n
Now, to prove p^2 = (ab)^n:
p^2 = [(ab)^n]^2
= (ab)^n * (ab)^n
= a^n * b^n * a^n * b^n
= (a^n * a^n) * (b^n * b^n)
= a^(2n) * b^(2n)
= (a^2 * b^2)^n
= (ab)^2n
Since both p^2 and (ab)^n reduce to (ab)^2n, we can conclude that p^2 = (ab)^n.
Therefore, the given equation p^2 = (ab)^n is proved for a geometric progression (G.P.).