Question

if the first and the nth term of GP for A and B respectively and if P is the product of n term prove that p square is equal to a b the whole power n​

Answer 1

Answer:

$\bold{P^2=(AB)^n}$

Step-by-step explanation:

Given:

first term,$t_1=a=A$

n th term,$t_n=ar^{n-1}=B$

Also

P=Product of first n terms

$P=a.ar.ar^2......ar^{n-1}$

$P=a^n.r^{1+2+3+....+n-1}$

$P=a^n.r^{\frac{n(n-1)}{2}}$

squaring on both sides, we get

$P^2=(a^n.r^{\frac{n(n-1)}{2}})^2$

$P^2=a^{2n}.r^{\frac{2n(n-1)}{2}}$

$P^2=a^{2n}.r^{n(n-1)}$

$P^2=(a^2.r^{n-1})^n$

$P^2=(a.ar^{n-1})^n$

$P^2=(a.ar^{n-1})^n$

$P^2=(A.B)^n$

$\implies\:\bold{P^2=(AB)^n}$