if a^n+b^n/a^n-2+b^n-2 is the A.M between a and b then find the value of n
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Answer:
We know that arithmetic mean between a and b is A.M=
2
a+b
Given: A.M between a and b is
a
n−1
+b
n−1
a
n
+b
n
So,
a
n−1
+b
n−1
a
n
+b
n
=
2
a+b
⇒2(a
n
+b
n
)=(a+b)(a
n−1
+b
n−1
)
⇒2a
n
+2b
n
=a(a
n−1
+b
n−1
)+b(a
n−1
+b
n−1
)
⇒2a
n
+2b
n
=a
n
+ab
n−1
+ba
n−1
+b
n
⇒2a
n
+2b
n
−a
n
−ab
n−1
−ba
n−1
−b
n
=0
⇒a
n
−ba
n−1
+b
n
−ab
n−1
=0
⇒a
n−1
(a−b)−b
n−1
(a−b)=0
a
n−1
−b
n−1
=0 or a−b=0
a
n−1
=b
n−1
or a=b but a
=b
∴
b
n−1
a
n−1
=1
⇒(
b
a
)
n−1
=(
b
a
)
0
since a
0
=1
Since bases are same, we can equate the powers,
∴n−1=0
Hence n=1
Step-by-step explanation:
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