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 P2 = (ab) n .
Let the given geometric progression (G.P.) have first term a and common ratio r.
Then the nth term of the G.P. is \[ ar^{\,n-1}. \]
It is given that the nth term is b. Hence, \[ ar^{\,n-1} = b \quad \text{… (1)} \]
The n terms of the G.P. are: \[ a,\; ar,\; ar^2,\; \ldots,\; ar^{n-1}. \]
Let P be the product of these n terms. Then,
\[ P = a \cdot ar \cdot ar^2 \cdot \ldots \cdot ar^{n-1}. \]
Taking a common from each term:
\[ P = a^n \cdot r^{\,0+1+2+\cdots+(n-1)}. \]
We know that \[ 0+1+2+\cdots+(n-1) = \frac{n(n-1)}{2}. \]
Therefore, \[ P = a^n r^{\frac{n(n-1)}{2}}. \]
Squaring both sides, we get \[ P^2 = a^{2n} r^{n(n-1)}. \]
From equation (1), we have \[ r^{n-1} = \frac{b}{a}. \]
Raising both sides to the power n: \[ r^{n(n-1)} = \left(\frac{b}{a}\right)^n. \]
Substituting this value in the expression for \( P^2 \):
\[ P^2 = a^{2n} \left(\frac{b}{a}\right)^n. \]
\[ P^2 = a^{2n} \cdot \frac{b^n}{a^n}. \]
\[ P^2 = a^n b^n. \]
\[ P^2 = (ab)^n. \]
Hence, it is proved that \[ P^2 = (ab)^n. \]