#### Question

If the first and the *n*^{th} term of a G.P. are *a* ad *b*, respectively, and if *P* is the product of *n *terms, prove that *P*^{2} = (*ab*)^{n}.

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#### Solution

The first term of the G.P is *a* and the last term is *b*.

Therefore, the G.P. is *a*, *ar*, *ar*^{2}, *ar*^{3}, … *ar*^{n}^{–1}, where *r* is the common ratio.

*b* = *ar*^{n}^{–1} … (1)

*P* = Product of *n* terms

= (*a*) (*ar*) (*ar*^{2}) … (*ar*^{n}^{–1})

= (*a* × *a* ×…*a*) (*r* × *r*^{2} × …*r*^{n}^{–1})

= *a*^{n}^{ }*r* ^{1 + 2 +…(}^{n}^{–1)} … (2)

Here, 1, 2, …(*n* – 1) is an A.P.

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Solution for question: 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. concept: Geometric Progression (G. P.). For the courses CBSE (Arts), CBSE (Commerce), CBSE (Science)