In a finite GP, prove that the product of the terms equidistant from t

1.	In a finite GP, prove that the product of the terms equidistant from the beginning and end is the product of first and last terms.
Answer» We need to prove that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP. Let us first consider a finite GP. A, AR, AR²….AR^{n -1} , ARⁿ . Where n is finite. Product of first and last terms in the given GP = A.ARⁿ = A²Rⁿ → (a) Now, n^th term of the GP from the beginning = AR^n-1 → (1) Now, starting from the end, First term = ARⁿ Last term = A \(\frac{1}{R} \) = Common Ratio So, an n^th term from the end of GP, A_n = (ARⁿ)\(\big(\frac{1}{R^{n-1}} \big)\) = AR → (2) So, the product of nth terms from the beginning and end of the considered GP from (1) and (2) = (AR^n-1 ) (AR) = A²Rⁿ → (b) So, from (a) and (b) its proved that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.

In a finite GP, prove that the product of the terms equidistant from the beginning and end is the product of first and last terms.

Answer»

We need to prove that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.

Let us first consider a finite GP.

A, AR, AR²….AR^{n -1} , ARⁿ .

Where n is finite. Product of first and last terms in the given GP = A.ARⁿ = A²Rⁿ → (a)

Now, n^th term of the GP from the beginning = AR^n-1 → (1)

Now, starting from the end,

First term = ARⁿ

Last term = A

\(\frac{1}{R} \) = Common Ratio

So, an n^th term from the end of GP, A_n = (ARⁿ)\(\big(\frac{1}{R^{n-1}} \big)\) = AR → (2)

So, the product of nth terms from the beginning and end of the considered GP from (1) and (2) = (AR^n-1 ) (AR)

= A²Rⁿ → (b)

So, from (a) and (b) its proved that the product of the terms equidistant from the beginning and end is the product of first and last terms in a finite GP.

In a finite GP, prove that the product of the terms equidistant from the beginning and end is the product of first and last terms.

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