InterviewSolution
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InterviewSolution
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If p is the first of the n arithmetic means between two numbers and q be the first on n harmonic means between the same numbers. Then, show that q does not lie between p and `((n+1)/(n-1))^2 p.`A. q lies between p and `((n+1)/(n-1))^(2)p`B. q lies between p and `((n+1)/(n-1))p`C. q does not lie between p and `((n+1)/(n-1))^(2)p`D. q does not lie between p and `((n+1)/(n-1))p` |
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Answer» Correct Answer - C For `ngt1`, we have `n+1gtn-1` `implies (n+1)/(n-1)gt1 implies p((n+1)/(n-1))^(2)gtp" " [:.pgt0]".......(i)"` Now, `p=a+d` Since,a,p,b are in AP. And `d=(b-a)/(n+1)` Again, ` (p)/(q)=((an+b)/(n+1))xx[(a+bn)/(ab(n+1))]` ` =(a^(2)n+abn^(2)+b^(2)n+ab)/(ab(n+1)^(2))` ` =(n((a)/(b)+(b)/(a))+(n^(2)+1))/((n+1)^(2))` `implies (p)/(q)-1=(n((a)/(b)+(b)/(a)-2))/((n+1)^(2))=(n(sqrt(a)/sqrt(b)+sqrt(b)/sqrt(a))^(2))/((n+1)^(2))` So, `(p)/(q)-1gt0 implies (p)/(q)gt1 implies pgtq" " "........(iii)"` From Eqs. (i) and (ii), we get `qltplt((n+1)/(n-1))^(2)p`. |
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