A Geometric Progression Problem That I Swear Will Not Fail To Confuse You

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)}.

(Photograph Source)

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 r² ) ............ 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) 
= p²ⁿ / 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.