# 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 P^{2} = (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, ar^{2}, ar^{3}, ...., ar^{n }^{- 1} and b = ar^{n }^{- }^{1}....(1)

Now, the product of n terms

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

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

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

= [a^{2}r^{(n - 1)}]^{n}

= [a x ar^{(n - 1)}]^{n} ....[Using (1)]

P^{2} = (ab)^{n}

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 P^{2} = (ab)^{n}

**Summary:**

The first and the nth term of the G.P were a and b, and we proved that P^{2} = (ab)^{n}