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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 61 of 106
Marks: +1, -0
Find the value of n so that an+1+bn+1an+bn\frac{a^{n+1}+b^{n+1}}{a^n+b^n} may be the geometric mean between a and b.
Solution:  
If an+1+bn+1an+bn\frac{a^{n+1}+b^{n+1}}{a^n+b^n} is G.M. between a and b, Then an+1+bn+1an+bn\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = ab\sqrt{ab}
⇒ an+1+bn+1a^{n+1}+b^{n+1} = (an+bn)ab(a^n+b^n)\sqrt{ab} ⇒ an+1+bn+1a^{n+1}+b^{n+1} = an+1/2b1/2+a1/2bn+1/2a^{n+1/2} b^{1/2} + a^{1/2} b^{n+1/2}
⇒ an+1−an+1/2b1/2a^{n+1} - a^{n+1/2} b^{1/2} = a1/2bn+1/2−bn+1a^{1/2} b^{n+1/2} - b^{n+1}
⇒ an+1/2(a1/2−b1/2)a^{n+1/2}(a^{1/2}-b^{1/2}) = a1/2bn+1/2−bn+1a^{1/2} b^{n+1/2} - b^{n+1}
⇒ an+1/2(a1/2−b1/2)a^{n+1/2} (a^{1/2}-b^{1/2}) = bn+1/2(a1/2−b1/2)b^{n+1/2} (a^{1/2}-b^{1/2})
⇒ an+1/2a^{n+1/2} = bn+1/2b^{n+1/2} ⇒ (ab)n+1/2\left(\frac{a}{b}\right)^{n+1/2} = 1 = (ab)0\left(\frac{a}{b}\right)^0 ⇒ n + 12\frac{1}{2} = 0 ⇒ n = −12-\frac{1}{2}
Hence the value of n is −12-\frac{1}{2}
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