English

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. Then prove that xb – c. yc – a . za – b = 1 - Mathematics

Advertisements
Advertisements

Question

If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. Then prove that xb – c. yc – a . za – b = 1

Sum

Solution

We have a, b, c as three consecutive terms of A.P.

Then b – a = c – b = d   ...(say)

c – a = 2d

a – b = – d

Now xb – c . yc – a . za – b = x–d . y2d . z–d

= `x^(-d) (sqrt(xz))^(2d) * z^(-d)`  ....`("Since"  y = (sqrt(xz)))  "as"  x, y, z  "are"  "G.P.")`

= `x^(-d) * x^d * z^d * z^(-d)`

= `x^(-d + d) * z^(d - d)`

= `x^o z^o` = 1

shaalaa.com
  Is there an error in this question or solution?
Chapter 9: Sequences and Series - Solved Examples [Page 156]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 9 Sequences and Series
Solved Examples | Q 12 | Page 156

RELATED QUESTIONS

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are `A+- sqrt((A+G)(A-G))`.


If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.


The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.


The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that `a:b = (m + sqrt(m^2 - n^2)):(m - sqrt(m^2 - n^2))`.


Find the A.M. between:

12 and −8


Find the A.M. between:

(x − y) and (x + y).


Insert 7 A.M.s between 2 and 17.


Insert six A.M.s between 15 and −13.


There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1. Find the value of n.


If a is the G.M. of 2 and \[\frac{1}{4}\] , find a.


Find the two numbers whose A.M. is 25 and GM is 20.


Construct a quadratic in x such that A.M. of its roots is A and G.M. is G.


If AM and GM of roots of a quadratic equation are 8 and 5 respectively, then obtain the quadratic equation.


If AM and GM of two positive numbers a and b are 10 and 8 respectively, find the numbers.


Prove that the product of n geometric means between two quantities is equal to the nth power of a geometric mean of those two quantities.


If the A.M. of two positive numbers a and b (a > b) is twice their geometric mean. Prove that:

\[a : b = (2 + \sqrt{3}) : (2 - \sqrt{3}) .\]


If A is the arithmetic mean and G1, G2 be two geometric means between any two numbers, then prove that 2A = `(G_1^2)/(G_2) + (G_2^2)/(G_1)`


The minimum value of 4x + 41–x, x ∈ R, is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×