If a, an+bn/an-1+bn-1 and b are in A.P and a≠b, then find the value of n.
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Since a, an+bn/an-1+bn-1 and b are in A.P, we have
(a + b)/2 = (an+bn /an-1+bn-1) [AM between two numbers in AP]
(a + b)( an-1+bn-1) = 2 an + 2bn
an + ban-1+ abn-1 + bn = 2an + 2bn
ban-1+ abn-1 = an + bn
abn-1 – bn = an – ban-1
bn-1( a – b) = an-1( a – b)
bn-1 = an-1
bn-1 / an-1 = 1
(b/a) n-1 = (b/a)0
On comparing both the sides, we get
n – 1 = 0
Therefore, n = 1
(a + b)/2 = (an+bn /an-1+bn-1) [AM between two numbers in AP]
(a + b)( an-1+bn-1) = 2 an + 2bn
an + ban-1+ abn-1 + bn = 2an + 2bn
ban-1+ abn-1 = an + bn
abn-1 – bn = an – ban-1
bn-1( a – b) = an-1( a – b)
bn-1 = an-1
bn-1 / an-1 = 1
(b/a) n-1 = (b/a)0
On comparing both the sides, we get
n – 1 = 0
Therefore, n = 1
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