Browse Questions

# If the first and $n^{th}$ term of a G.P. are $a$ and $b$ respectively and if $P$ is the product of the $n$ terms then prove that $P^2=(ab)^n$

Toolbox:
• $t_n=a.r^{n-1}$
• General G.P is $a,ar,ar^2,................ar^{n-1}$
• $1+2+3+.........n=\large\frac{n(n+1)}{2}$
• $(a^m)^n=a^{mn}$
• $a^m\times a^n=a^{m+n}$
Given that first term of a G.P=$a$ and $n^{th}$ term = $t_n=b$
$\Rightarrow\:b=a.r^{n-1}$
General G.P. is $a,ar,ar^2,ar^3,.............ar^{n-1}$
It is also given that the product of $n$ terms of a G.P $=P$
$i.e.,\:a\times (ar)\times (ar^2)\times (ar^3).............\times ((ar^{n-1})=P$
$\Rightarrow\:(a\times a\times a..........(n\:times)).(r\times r^2\times r^3\times........r^{n-1})$
We know that $a^m\times a^n\:\times a^p\times......=a^{m+n+p+.....}$
$\therefore\:(r\times r^2\times r^3\times........r^{n-1})=r^{1+2+3+......(n-1)}$
$\Rightarrow\:a^n.(r^{1+2+3+......+(n-1)})=P$
We know that $1+2+3+......n=\large\frac{n(n+1)}{2}$
$\therefore\:1+2+3+........(n-1)=\large\frac{(n-1)(n-1+1)}{2}=\frac{(n-1)n}{2}$
$\Rightarrow\:P=a^n.r^{\large\frac{(n-1)n}{2}}$
$\Rightarrow\:P^2=\bigg(a^n.r^{\large\frac{(n-1)n}{2}}\bigg)^2$
We know that $(a^m)^n=a^{mn}$
$\therefore\:P^2=\bigg(a^n.r^{\large\frac{(n-1)n}{2}}\bigg)^2=a^{2n}.r^{\large\frac{(n-1)n}{2}\times 2}$
$\Rightarrow\:P^2=a^{2n}.r^{(n-1)n}$........(i)
Step 2
Substituting the value of $b$ we get
$(ab)=a.(a.r^{n-1})=a^2.r^{n-1}$
$\therefore\:(ab)^n=(a^2.r^{n-1})^n=a^{2n}\:.\:r^{(n-1)n}$....(ii)
From (i) and (ii) we can say that $P^2=(ab)^n$
Hence proved.