If the first and n^th term of the G.P is a and b respectively, if P is the product of n terms, prove that P² = (ab)ⁿ

Solution:

It is given that the first term of the G.P is a and the last term is b .

Let the common ratio be r

Hence, the G.P is

a, ar, ar², ar³, ...., ar^{n - 1} and b = ar^{n - 1}....(1)

Now, the product of n terms

P = (a) x (ar) x (ar²) x .... x (ar^{n - 1})

= (a x a x a .... n times) (r x r² x .... x r^{n - 1})

= anr^{1 + 2 + .... + (n - 1) ....(2)}

Here, 1, 2, ...., (n - 1) is an A.P.

1+ 2 + .... + (n - 1) = (n - 1)/2 [2 + (n - 1 - 1) x 1]

= (n - 1)/2 [2 + n - 2]

= n (n - 1)/2

Substituting this value in (2) , we obtain

P = aⁿr^{n (n - 1)/2}

P² = a²ⁿr^{n(n - 1)}

= [a²r^{(n - 1)}]ⁿ

= [a x ar^{(n - 1)}]ⁿ ....[Using (1)]

P² = (ab)ⁿ

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NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 23

If the first and n^th term of the G.P is a and b respectively, if P is the product of n terms, prove that P² = (ab)ⁿ

Summary:

The first and the nth term of the G.P were a and b, and we proved that P² = (ab)ⁿ