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Let p(gt0) be the first of the n arthimaticmeans betweens between two numbers and q(gt0) the first of n harmonic means between the same numbers. Then prove that qnotin(p,((n+1)/(n-1))^(2)p)andpnotin(((n-1)/(n+1))^(2)q,q) |
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Answer» SOLUTION :Let a and b be two numbers and `A_(1),A_(2),A_(3),…A_(n)` be n A.M's between a and b. Then `a,A_(1),A_(2),…,A_(n)`,b are in A.P. There are (n+2) terms in the series `rArra+(n+1)d=b` `rArrd=(b-1)/(n+1)` `rArrA_(1)=p=a+(b-1)/(n+1)=(an+b)/(n+1)` ..(1) The first H.M between a and b, when nHM's are inserted between a and b can be obtained by replacing a by `1/a` and b by `1/b` in eq. (1) and then taking its reciprocal. Therefore,`q=1/(((1/a)n+1/b)/(n+1))=((n+1)AB)/(bn+a)` Substitute b=p(n+1)-an[from (1)] in equation (2) to get `aq+nq[p(n+1)-an]=(n+1)a[p(n+1)-an]` `rArrna^(2)-[(n+1)p+(n-1)q]a+npq=0` `rArr" Discriminant" ge0 (because ` is a real) `rArr[(n+1)p+(n-1)q]^(2)-4n^(2)pqge0` `rArr(n-1)^(2)q^(2)+{2(n^(2)-1)-4n^(2)}pq+(n+1)^(2)p^(2)ge0` `rArrq^(2)-2(n^(2)+1)/((n-1)^(2))pq+((n+1)/(n-1))^(2)p^(2)ge0` `rArr[q-p((n+1)/(n-1))^(2)][q-p]ge0` `rArrq` can not lie between p and q `((n+1)/(n-1))^(2)` ALSO `[p((n+1)/(n-1))^(2)-q][p-q]ge0` `rArr[p-q((n-1)/(n+1))^(2)][p-q]ge0` `rArr p` can not lie between q `((n-1)/(n+1))^(2)` and q |
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