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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 63 of 106
Marks: +1, -0
If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ± (A+G)(AG)\sqrt{(A+G)(A-G)}.
Solution:  
Let a and b be the numbers such that A, G are A.M. and G.M. respectively between them.
A.M. of a and b = a+b2\frac{a+b}{2} i.e., A = a+b2\frac{a+b}{2} ⇒ a + b = 2A ... (i)
Also, G.M. of a and b = ab\sqrt{ab} i.e., G = ab\sqrt{ab}
G2G^2 = ab .....(ii)
We know that, (ab)2(a - b)^2 = (a+b)2(a + b)^2 – 4ab
(ab)2(a - b)^2 = (2A)24G2(2A)^2 - 4G^2 [By using (i) and (ii)]
(ab)2(a - b)^2 = 4A24G24A^2 - 4G^2
⇒ a - b = 2 A2G2\sqrt{A^2 - G^2} ... (iii)
Adding (i) & (iii), we get
2a = 2A + 2A2G22\sqrt{A^2 - G^2} ⇒ a = A + (AG)(A+G)\sqrt{(A-G)(A+G)}
Subtracting (iii) from (i), we get
2b = 2A - 2 A2G2\sqrt{A^2 - G^2} ⇒ b = A - (AG)(A+G)\sqrt{(A-G)(A+G)}
Hence, the numbers are A ± (AG)(A+G)\sqrt{(A-G)(A+G)}
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