Question 23

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}.

Answer

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.

∴1 + 2 + ……….+ (n – 1)

Thus, the given result is proved.

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(iv) D = {x: x is a prime number which is divisor of 60}.

(v) E = The set of all letters in the word TRIGONOMETRY.

(vi) F = The set of all letters in the word BETTER. - Q:-
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Show that the sum of (

*m*+*n*)th and (*m*–*n*)th terms of an A.P. is equal to twice the*m**t*h term.

Nafeesa
2019-10-30 19:06:29

Thank you

Aniska Chatterjee
2019-07-28 08:25:27

Very good.. thanks for the easiest way.

Hitesh Kumar
2019-05-04 15:52:24

Very good

Yash
2019-02-09 12:56:13

Please change the colour of this page And nice explanation

Pranav
2019-02-06 23:30:16

Why n-1/2 is taken instead of n/2

Ravi Ranjan
2018-09-02 00:54:54

Change the colour of the page

Param
2018-06-27 23:56:10

Good

- NCERT Chapter