If the first and the nth term of a G.p are a and b,respectively ,and if P is the priduct of n terms,prove that P2=(ab)n.
106nth term of a GP = arn-1 where r=common ratio, a=first term
so. arn-1 = b ... (i)
product of n terms = a*ar*ar2....*arn-1
= an*r1+2+...+n-1
= an*rn(n-1)/2 = P
so, P2 = a2n*rn(n-1) ... (ii)
(ab)n = an*(arn-1))n = a2n*rn(n-1) ... (iii)
comparing (ii) and (iii), P2 = (ab)n
6but i think there is a mistake....
we know that 1+2+3...N=N(N-1)/2
and in this question we have ...r1+2+3+...n-1 (so here N=n-1)
=rn-1(n-2)/2 (and not =rn(n-1)/2).
Plz tell me wether im correct or not.
1357nix
1+2+3...N=N(N+1)/2 not N(N-1)/2
for N=N-1 it becomes N(N-1)/2