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NCERT Class XI Mathematics - Sequences and Series - Solutions

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Question : 29 of 106
Marks: +1, -0
If an+bnan−1+bn−1\frac{a^n+bn}{a^{n-1}+b^{n-1}} is the A.M. between a and b, then find the value of n.
Solution:  
We have an+bnan−1+bn−1\frac{a^n+bn}{a^{n-1}+b^{n-1}} = a+b2\frac{a+b}{2}
⇒ 2an+2b22a^n+2b^2 = an+abn−1+an−1b+bna^n+ab^{n-1}+a^{n-1}b+b^n ⇒ an+bna^n+b^n = abn−1+an−1bab^{n-1}+a^{n-1}b
⇒ an−abn−1+bn−an−1ba^n-ab^{n-1}+b^n-a^{n-1}b = 0 ⇒ a(an−1−bn−1)a(a^{n-1}-b^{n-1}) + b(bn−1−an−1)b(b^{n-1}-a^{n-1}) = 0
⇒ (a - b) (an−1−bn−1)(a^{n-1}-b^{n-1}) = 0 ⇒ an−1−bn−1a^{n-1}-b^{n-1} = 0 (Since a - b ≠ 0)
⇒ an−1a^{n-1} = bn−1b^{n-1} ⇒ an−1bn−1\frac{a^{n-1}}{b^{n-1}} = 1 ⇒ (a/b)n−1(a/b)^{n-1} = 1 = (a/b)0(a/b)^0
⇒ n - 1 = 0 ⇒ n = 1
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