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an+1+bn+1/an+bn\xa0is the A.M between a and b, then find the value of n.\xa0 |
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Answer» Ans.\xa0As we knowA.M. of two numbers a and b = (a+b)/2So(an+1\xa0+ bn+1)/(an+bn) = (a+b)/2=> 2(an+1+bn+1) = (a+b)(an+bn)=> 2an+1\xa0+ 2bn+1\xa0= an+1\xa0+ abn\xa0+ ban\xa0+ bn+1=> an+1\xa0- ban\xa0+ bn+1\xa0- abn\xa0= 0=> an(a-b) + bn( b-a) = 0=> an(a-b) -\xa0bn(a-b) = 0=> an(a-b) = bn(a-b)=> an/bn\xa0= 1=> (a/b)n\xa0= (a/b)0=> n =0value of n = 0 According to question,an+1+bn+1/an+bn\xa0= (a + b) / 2=> 2(an+1 + bn+1) = (a + b)(an + bn)=> 2an+1 + 2bn+1 =\xa0an+1 + abn + anb + bn+1=> an+1\xa0-\xa0anb +\xa0bn+1 - abn = 0=> an\xa0(a -\xa0b) - bn (a - b)= 0=> (an\xa0-\xa0bn)(a- b)= 0=> (an\xa0-\xa0bn) = 0=> an\xa0= bn=> (an/bn) = 1=> (a/b)n = (a/b)0=> n = 0\xa0\xa0\xa0 |
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