If A and G be A.M and G.M, respectively between two positive numbers, prove that the numbers are A ± √( A + G)( A - G
Solution:
It is given that A and G are A.M and G.M, respectively between two positive numbers.
Let these two positive numbers be a and b.
Therefore,
A.M = A = (a + b) / 2
⇒ a + b = 2 A ....(1)
And,
G.M = G = √ab
⇒ ab = G2 ....(2)
Since,
(a - b)2 = (a + b)2 - 4ab
= 4 A2 - 4G2 [Using (1) and (2)]
= 4( A2 - G2)
= 4( A + G)( A - G)
(a - b) = 2√( A + G)( A - G) ....(3)
By adding (1) and (3), we obtain
⇒ 2a = 2A + 2√( A + G)( A - G)
⇒ a = A + √( A + G)( A - G)
Substituting the value of a in (1) , we obtain
b = 2 A - (A + √( A + G)( A - G))
= A - √( A + G)( A - G))
Thus, the two numbers are A ± √( A + G)( A - G))
NCERT Solutions Class 11 Maths Chapter 9 Exercise 9.3 Question 29
If A and G be A.M and G.M, respectively between two positive numbers, prove that the numbers are A ± √( A + G)( A - G)
Summary:
A and G are the A.M and G.M, respectively between two positive numbers we have proved that the numbers are A ± √( A + G)( A - G))
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