**Question : If the n**

^{th}term of a G.P is p,then show that the product of the first (2n - 1) terms is p^{(2n - 1)}.

**I figured out this question when I got assigned to solve it in Maths class at school in the Chhaya Prakashani Class XI Maths book written by Sourendranath De.Nonetheless, let me get straight into the problem.**

**So,it's almost like,what?Multiplication of terms that are in G.P? How to solve this?As I bwgun attempting to solve the problem,I observed that plain statistics would work and if you were still in doubts,you could check if you were right by a back calculation process.Here's how.**

__Solution__

**According to the question,we have a . r**

^{(n-1)}= p ...... (i)**Now,the (2p - 1)**

^{th}term = a . r^{(2n -1 -1)}= a.r^{(2n -2)}...... (ii)

**Now,let us multiply from the first term of the G.P up to the**

**(2p - 1)**

^{th}term.**(a) x (a x r) ( a x r**

^{2}) ............**a . r**

^{(n-1)}.............**a.r**

^{(2n -2)}**= a**

^{(2n - 1)}x r^{{1 + 2 + 3 + ...... + (2n - 2)}}**Now,notice that the exponents of r are in A.P.Therefore,**

**= a**

^{(2n - 1)}x r^{{(2n - 2)/2(1 + 2n - 2)}}

^{= }**a**

^{(2n - 1)}x r^{{(n - 1)( 2n - 1)}}

^{= }**a**

^{(2n - 1)}x r^{(2n2 - n - 2n + 1 )}

^{= }**a**

^{(2n - 1)}x r^{(2n2 - 3n + 1 )}**= a**

^{(2n)}x r^{2n(n - 1)}/**a . r**

^{(n-1)}**=**

**a . r**

^{(n-1)2n}/**a . r**

^{(n-1)}**= p**

^{2n}/ p ....... (From Eqn 1)**=**

**p**

^{(2n - 1)}

^{}**Now,you might calculate what value you get by**

**p**

^{(2n - 1)}**,which I referred to as the back caluclation process.Hope this helps.**

**Post Script - I am suffering from Dengue fever.Sighs.But still trying my best to keep up eith everything else.**

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