The given G.P. is 1, – a, a2, – a3, ......
Here, first term = a1 = 1
Common ratio = r = a
Sn = \(\frac{a^1(1-r^n)}{1-r}\)
∴Sn = \(\frac{1[1-(-a)^n]}{1-(-a)}\)
= \(\frac{[1-(-a)n]}{1+a}\)
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 .