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

Asked by Pragya Singh | 1 year ago |  84

##### Solution :-

Given, the first term of the G.P is a and the last term is b.

Thus,

The G.P. is a, ar, ar2, ar3, … arn–1, where r is the common ratio.

Then,

b = arn–1  … (1)

P = Product of n terms

= (a) (ar) (ar2) … (arn–1)

= (a × a ×…a) (r × r2 × …rn–1)

= a1 + 2 +…(n–1)  … (2)

Here, 1, 2, …(n – 1) is an A.P.

$$\dfrac{n-1}{2}[2+(n-1-1)\times 1]$$

$$\dfrac{n-1}{2}[2+n-2]$$

$$\dfrac{n(n-1)}{2}$$

And, the product of n terms P is given by,

$$p=a^nr^\dfrac{n(n-1)}{2}$$

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

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

$$[a\times ar^{n-1}]^n$$

Substituting (1) in the equation,

p2= (ab)n

Therefore, p2= (ab)is proved.

Answered by Abhisek | 1 year ago

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