118687
If \(H\) is harmonic mean between \(P\) and \(Q\). Then the value of \(\frac{H}{P}+\frac{H}{Q}\) is
1 2
2 \(\frac{P Q}{P+Q}\)
3 \(\frac{\mathrm{P}+\mathrm{Q}}{\mathrm{PQ}}\)
4 None of these
Explanation:
A Given, Harmonic mean of \(\mathrm{P}\) and \(\mathrm{Q}\) is \(\mathrm{H}\). \(\text { So, } \frac{2}{\mathrm{H}}=\frac{1}{\mathrm{P}}+\frac{1}{\mathrm{Q}}\) \(\frac{2 \mathrm{PQ}}{\mathrm{H}}=(\mathrm{P}+\mathrm{Q})\) \(\mathrm{H}=\frac{2 \mathrm{PQ}}{\mathrm{P}+\mathrm{Q}}\) On dividing by \(\mathrm{P}\) on both side of equation (i), we get, \(\frac{\mathrm{H}}{\mathrm{P}}=\frac{2 \mathrm{Q}}{(\mathrm{P}+\mathrm{Q})}\) Similarly- \(\frac{H}{Q}=\frac{2 P}{(P+Q)}\) On adding equation (ii) and equation (iii), we get - \(\frac{H}{P}+\frac{H}{Q}=\frac{2 P+2 Q}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=\frac{2(P+Q)}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=2\)
UPSEE-2010
Sequence and Series
118688
If positive numbers \(a, b, c\) are in \(H P\) and \(c>a\), then \(\log (a+c)+\log (a-2 b+c)\) is equal to
1 \(2 \log (c-b)\)
2 \(2 \log (a+c)\)
3 \(2 \log (\mathrm{c}-\mathrm{a})\)
4 \(2 \log (a-c)\)
Explanation:
C Given, numbers \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are in H.P. So, harmonic mean of these numbers - \(\mathrm{b}=\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}\) Let, \(\mathrm{x}=\log (\mathrm{a}+\mathrm{c})+\log (\mathrm{a}-2 \mathrm{~b}+\mathrm{c})\) On putting the value of \(b\) in equation (i) we get - \(x=\log (a+c)+\log \left(a+c-2\left(\frac{2 a c}{a+c}\right)\right)\) \(x=\log (a+c)+\log \left[\frac{(a+c)^2-4 a c}{(a+c)}\right]\) \(x=\log (a+c)+\log \left[\frac{\left(a^2+c^2+2 a c-4 a c\right)}{a+c}\right]\) \(x=\log (a+c)+\log \left[\frac{(c-a)^2}{(a+c)}\right] \quad[\therefore c>a]\) \(x=\log \left[(a+c) \times \frac{(c-a)^2}{(a+c)}\right]\) \(x=2 \log (c-a) .\)
UPSEE-2009
Sequence and Series
118689
If the AM of two numbers be \(A\) and GM be G, then the numbers will be
C Let the numbers are \(\mathrm{a}\) and \(\mathrm{b}\). Arithmetic mean of \(a\) and \(b\). \(A=\frac{a+b}{2} \Rightarrow a+b=2 A\) And geometric mean of \(a\) and \(b\) \(\mathrm{G}=\sqrt{\mathrm{ab}}\) \(\mathrm{G}^2=\mathrm{ab}\) Now, \((a-b)=\sqrt{(a+b)^2-4 a b}\) From equation (i), (ii) and (iii) we get - \((a-b)=\sqrt{4 A^2-4 G^2}\) \(a-b=2 \sqrt{(A-G)(A+G)}\) Solving equation (i) and (iv) we get \(a=A+\sqrt{(A-G)(A+G)}\) On putting the value a in equation (i) \(\mathrm{b}=\mathrm{A}-\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\) So, the numbers are \(A \pm \sqrt{(A-G)(A+G)}\)
UPSEE -2008
Sequence and Series
118690
If arithmetic mean of two positive numbers is \(A\), their geometric mean is \(G\) and harmonic mean is \(H\), then \(H\) is equal to:
1 \(G^2 / A\)
2 \(\mathrm{A}^2 / \mathrm{G}^2\)
3 \(\mathrm{A} / \mathrm{G}^2\)
4 \(\mathrm{G} / \mathrm{A}^2\)
Explanation:
A Let the numbers are \(a\) and \(b\). Then, arithmetic mean \((\mathrm{A})=\frac{\mathrm{a}+\mathrm{b}}{2}\) \(\mathrm{a}+\mathrm{b}=2 \mathrm{~A}\) Geometric mean, \(\mathrm{G}=\sqrt{\mathrm{ab}}\) or \(\mathrm{G}^2=\mathrm{ab}\) Harmonic mean, \(\mathrm{H}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}\) On putting the value of equation (i) and (ii) in equation (iii), \(\therefore \mathrm{H}=\frac{2 \mathrm{G}^2}{2 \mathrm{~A}}\) \(\mathrm{H}=\frac{\mathrm{G}^2}{\mathrm{~A}}\)
118687
If \(H\) is harmonic mean between \(P\) and \(Q\). Then the value of \(\frac{H}{P}+\frac{H}{Q}\) is
1 2
2 \(\frac{P Q}{P+Q}\)
3 \(\frac{\mathrm{P}+\mathrm{Q}}{\mathrm{PQ}}\)
4 None of these
Explanation:
A Given, Harmonic mean of \(\mathrm{P}\) and \(\mathrm{Q}\) is \(\mathrm{H}\). \(\text { So, } \frac{2}{\mathrm{H}}=\frac{1}{\mathrm{P}}+\frac{1}{\mathrm{Q}}\) \(\frac{2 \mathrm{PQ}}{\mathrm{H}}=(\mathrm{P}+\mathrm{Q})\) \(\mathrm{H}=\frac{2 \mathrm{PQ}}{\mathrm{P}+\mathrm{Q}}\) On dividing by \(\mathrm{P}\) on both side of equation (i), we get, \(\frac{\mathrm{H}}{\mathrm{P}}=\frac{2 \mathrm{Q}}{(\mathrm{P}+\mathrm{Q})}\) Similarly- \(\frac{H}{Q}=\frac{2 P}{(P+Q)}\) On adding equation (ii) and equation (iii), we get - \(\frac{H}{P}+\frac{H}{Q}=\frac{2 P+2 Q}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=\frac{2(P+Q)}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=2\)
UPSEE-2010
Sequence and Series
118688
If positive numbers \(a, b, c\) are in \(H P\) and \(c>a\), then \(\log (a+c)+\log (a-2 b+c)\) is equal to
1 \(2 \log (c-b)\)
2 \(2 \log (a+c)\)
3 \(2 \log (\mathrm{c}-\mathrm{a})\)
4 \(2 \log (a-c)\)
Explanation:
C Given, numbers \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are in H.P. So, harmonic mean of these numbers - \(\mathrm{b}=\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}\) Let, \(\mathrm{x}=\log (\mathrm{a}+\mathrm{c})+\log (\mathrm{a}-2 \mathrm{~b}+\mathrm{c})\) On putting the value of \(b\) in equation (i) we get - \(x=\log (a+c)+\log \left(a+c-2\left(\frac{2 a c}{a+c}\right)\right)\) \(x=\log (a+c)+\log \left[\frac{(a+c)^2-4 a c}{(a+c)}\right]\) \(x=\log (a+c)+\log \left[\frac{\left(a^2+c^2+2 a c-4 a c\right)}{a+c}\right]\) \(x=\log (a+c)+\log \left[\frac{(c-a)^2}{(a+c)}\right] \quad[\therefore c>a]\) \(x=\log \left[(a+c) \times \frac{(c-a)^2}{(a+c)}\right]\) \(x=2 \log (c-a) .\)
UPSEE-2009
Sequence and Series
118689
If the AM of two numbers be \(A\) and GM be G, then the numbers will be
C Let the numbers are \(\mathrm{a}\) and \(\mathrm{b}\). Arithmetic mean of \(a\) and \(b\). \(A=\frac{a+b}{2} \Rightarrow a+b=2 A\) And geometric mean of \(a\) and \(b\) \(\mathrm{G}=\sqrt{\mathrm{ab}}\) \(\mathrm{G}^2=\mathrm{ab}\) Now, \((a-b)=\sqrt{(a+b)^2-4 a b}\) From equation (i), (ii) and (iii) we get - \((a-b)=\sqrt{4 A^2-4 G^2}\) \(a-b=2 \sqrt{(A-G)(A+G)}\) Solving equation (i) and (iv) we get \(a=A+\sqrt{(A-G)(A+G)}\) On putting the value a in equation (i) \(\mathrm{b}=\mathrm{A}-\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\) So, the numbers are \(A \pm \sqrt{(A-G)(A+G)}\)
UPSEE -2008
Sequence and Series
118690
If arithmetic mean of two positive numbers is \(A\), their geometric mean is \(G\) and harmonic mean is \(H\), then \(H\) is equal to:
1 \(G^2 / A\)
2 \(\mathrm{A}^2 / \mathrm{G}^2\)
3 \(\mathrm{A} / \mathrm{G}^2\)
4 \(\mathrm{G} / \mathrm{A}^2\)
Explanation:
A Let the numbers are \(a\) and \(b\). Then, arithmetic mean \((\mathrm{A})=\frac{\mathrm{a}+\mathrm{b}}{2}\) \(\mathrm{a}+\mathrm{b}=2 \mathrm{~A}\) Geometric mean, \(\mathrm{G}=\sqrt{\mathrm{ab}}\) or \(\mathrm{G}^2=\mathrm{ab}\) Harmonic mean, \(\mathrm{H}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}\) On putting the value of equation (i) and (ii) in equation (iii), \(\therefore \mathrm{H}=\frac{2 \mathrm{G}^2}{2 \mathrm{~A}}\) \(\mathrm{H}=\frac{\mathrm{G}^2}{\mathrm{~A}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Sequence and Series
118687
If \(H\) is harmonic mean between \(P\) and \(Q\). Then the value of \(\frac{H}{P}+\frac{H}{Q}\) is
1 2
2 \(\frac{P Q}{P+Q}\)
3 \(\frac{\mathrm{P}+\mathrm{Q}}{\mathrm{PQ}}\)
4 None of these
Explanation:
A Given, Harmonic mean of \(\mathrm{P}\) and \(\mathrm{Q}\) is \(\mathrm{H}\). \(\text { So, } \frac{2}{\mathrm{H}}=\frac{1}{\mathrm{P}}+\frac{1}{\mathrm{Q}}\) \(\frac{2 \mathrm{PQ}}{\mathrm{H}}=(\mathrm{P}+\mathrm{Q})\) \(\mathrm{H}=\frac{2 \mathrm{PQ}}{\mathrm{P}+\mathrm{Q}}\) On dividing by \(\mathrm{P}\) on both side of equation (i), we get, \(\frac{\mathrm{H}}{\mathrm{P}}=\frac{2 \mathrm{Q}}{(\mathrm{P}+\mathrm{Q})}\) Similarly- \(\frac{H}{Q}=\frac{2 P}{(P+Q)}\) On adding equation (ii) and equation (iii), we get - \(\frac{H}{P}+\frac{H}{Q}=\frac{2 P+2 Q}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=\frac{2(P+Q)}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=2\)
UPSEE-2010
Sequence and Series
118688
If positive numbers \(a, b, c\) are in \(H P\) and \(c>a\), then \(\log (a+c)+\log (a-2 b+c)\) is equal to
1 \(2 \log (c-b)\)
2 \(2 \log (a+c)\)
3 \(2 \log (\mathrm{c}-\mathrm{a})\)
4 \(2 \log (a-c)\)
Explanation:
C Given, numbers \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are in H.P. So, harmonic mean of these numbers - \(\mathrm{b}=\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}\) Let, \(\mathrm{x}=\log (\mathrm{a}+\mathrm{c})+\log (\mathrm{a}-2 \mathrm{~b}+\mathrm{c})\) On putting the value of \(b\) in equation (i) we get - \(x=\log (a+c)+\log \left(a+c-2\left(\frac{2 a c}{a+c}\right)\right)\) \(x=\log (a+c)+\log \left[\frac{(a+c)^2-4 a c}{(a+c)}\right]\) \(x=\log (a+c)+\log \left[\frac{\left(a^2+c^2+2 a c-4 a c\right)}{a+c}\right]\) \(x=\log (a+c)+\log \left[\frac{(c-a)^2}{(a+c)}\right] \quad[\therefore c>a]\) \(x=\log \left[(a+c) \times \frac{(c-a)^2}{(a+c)}\right]\) \(x=2 \log (c-a) .\)
UPSEE-2009
Sequence and Series
118689
If the AM of two numbers be \(A\) and GM be G, then the numbers will be
C Let the numbers are \(\mathrm{a}\) and \(\mathrm{b}\). Arithmetic mean of \(a\) and \(b\). \(A=\frac{a+b}{2} \Rightarrow a+b=2 A\) And geometric mean of \(a\) and \(b\) \(\mathrm{G}=\sqrt{\mathrm{ab}}\) \(\mathrm{G}^2=\mathrm{ab}\) Now, \((a-b)=\sqrt{(a+b)^2-4 a b}\) From equation (i), (ii) and (iii) we get - \((a-b)=\sqrt{4 A^2-4 G^2}\) \(a-b=2 \sqrt{(A-G)(A+G)}\) Solving equation (i) and (iv) we get \(a=A+\sqrt{(A-G)(A+G)}\) On putting the value a in equation (i) \(\mathrm{b}=\mathrm{A}-\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\) So, the numbers are \(A \pm \sqrt{(A-G)(A+G)}\)
UPSEE -2008
Sequence and Series
118690
If arithmetic mean of two positive numbers is \(A\), their geometric mean is \(G\) and harmonic mean is \(H\), then \(H\) is equal to:
1 \(G^2 / A\)
2 \(\mathrm{A}^2 / \mathrm{G}^2\)
3 \(\mathrm{A} / \mathrm{G}^2\)
4 \(\mathrm{G} / \mathrm{A}^2\)
Explanation:
A Let the numbers are \(a\) and \(b\). Then, arithmetic mean \((\mathrm{A})=\frac{\mathrm{a}+\mathrm{b}}{2}\) \(\mathrm{a}+\mathrm{b}=2 \mathrm{~A}\) Geometric mean, \(\mathrm{G}=\sqrt{\mathrm{ab}}\) or \(\mathrm{G}^2=\mathrm{ab}\) Harmonic mean, \(\mathrm{H}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}\) On putting the value of equation (i) and (ii) in equation (iii), \(\therefore \mathrm{H}=\frac{2 \mathrm{G}^2}{2 \mathrm{~A}}\) \(\mathrm{H}=\frac{\mathrm{G}^2}{\mathrm{~A}}\)
118687
If \(H\) is harmonic mean between \(P\) and \(Q\). Then the value of \(\frac{H}{P}+\frac{H}{Q}\) is
1 2
2 \(\frac{P Q}{P+Q}\)
3 \(\frac{\mathrm{P}+\mathrm{Q}}{\mathrm{PQ}}\)
4 None of these
Explanation:
A Given, Harmonic mean of \(\mathrm{P}\) and \(\mathrm{Q}\) is \(\mathrm{H}\). \(\text { So, } \frac{2}{\mathrm{H}}=\frac{1}{\mathrm{P}}+\frac{1}{\mathrm{Q}}\) \(\frac{2 \mathrm{PQ}}{\mathrm{H}}=(\mathrm{P}+\mathrm{Q})\) \(\mathrm{H}=\frac{2 \mathrm{PQ}}{\mathrm{P}+\mathrm{Q}}\) On dividing by \(\mathrm{P}\) on both side of equation (i), we get, \(\frac{\mathrm{H}}{\mathrm{P}}=\frac{2 \mathrm{Q}}{(\mathrm{P}+\mathrm{Q})}\) Similarly- \(\frac{H}{Q}=\frac{2 P}{(P+Q)}\) On adding equation (ii) and equation (iii), we get - \(\frac{H}{P}+\frac{H}{Q}=\frac{2 P+2 Q}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=\frac{2(P+Q)}{(P+Q)}\) \(\frac{H}{P}+\frac{H}{Q}=2\)
UPSEE-2010
Sequence and Series
118688
If positive numbers \(a, b, c\) are in \(H P\) and \(c>a\), then \(\log (a+c)+\log (a-2 b+c)\) is equal to
1 \(2 \log (c-b)\)
2 \(2 \log (a+c)\)
3 \(2 \log (\mathrm{c}-\mathrm{a})\)
4 \(2 \log (a-c)\)
Explanation:
C Given, numbers \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are in H.P. So, harmonic mean of these numbers - \(\mathrm{b}=\frac{2 \mathrm{ac}}{\mathrm{a}+\mathrm{c}}\) Let, \(\mathrm{x}=\log (\mathrm{a}+\mathrm{c})+\log (\mathrm{a}-2 \mathrm{~b}+\mathrm{c})\) On putting the value of \(b\) in equation (i) we get - \(x=\log (a+c)+\log \left(a+c-2\left(\frac{2 a c}{a+c}\right)\right)\) \(x=\log (a+c)+\log \left[\frac{(a+c)^2-4 a c}{(a+c)}\right]\) \(x=\log (a+c)+\log \left[\frac{\left(a^2+c^2+2 a c-4 a c\right)}{a+c}\right]\) \(x=\log (a+c)+\log \left[\frac{(c-a)^2}{(a+c)}\right] \quad[\therefore c>a]\) \(x=\log \left[(a+c) \times \frac{(c-a)^2}{(a+c)}\right]\) \(x=2 \log (c-a) .\)
UPSEE-2009
Sequence and Series
118689
If the AM of two numbers be \(A\) and GM be G, then the numbers will be
C Let the numbers are \(\mathrm{a}\) and \(\mathrm{b}\). Arithmetic mean of \(a\) and \(b\). \(A=\frac{a+b}{2} \Rightarrow a+b=2 A\) And geometric mean of \(a\) and \(b\) \(\mathrm{G}=\sqrt{\mathrm{ab}}\) \(\mathrm{G}^2=\mathrm{ab}\) Now, \((a-b)=\sqrt{(a+b)^2-4 a b}\) From equation (i), (ii) and (iii) we get - \((a-b)=\sqrt{4 A^2-4 G^2}\) \(a-b=2 \sqrt{(A-G)(A+G)}\) Solving equation (i) and (iv) we get \(a=A+\sqrt{(A-G)(A+G)}\) On putting the value a in equation (i) \(\mathrm{b}=\mathrm{A}-\sqrt{(\mathrm{A}-\mathrm{G})(\mathrm{A}+\mathrm{G})}\) So, the numbers are \(A \pm \sqrt{(A-G)(A+G)}\)
UPSEE -2008
Sequence and Series
118690
If arithmetic mean of two positive numbers is \(A\), their geometric mean is \(G\) and harmonic mean is \(H\), then \(H\) is equal to:
1 \(G^2 / A\)
2 \(\mathrm{A}^2 / \mathrm{G}^2\)
3 \(\mathrm{A} / \mathrm{G}^2\)
4 \(\mathrm{G} / \mathrm{A}^2\)
Explanation:
A Let the numbers are \(a\) and \(b\). Then, arithmetic mean \((\mathrm{A})=\frac{\mathrm{a}+\mathrm{b}}{2}\) \(\mathrm{a}+\mathrm{b}=2 \mathrm{~A}\) Geometric mean, \(\mathrm{G}=\sqrt{\mathrm{ab}}\) or \(\mathrm{G}^2=\mathrm{ab}\) Harmonic mean, \(\mathrm{H}=\frac{2 \mathrm{ab}}{\mathrm{a}+\mathrm{b}}\) On putting the value of equation (i) and (ii) in equation (iii), \(\therefore \mathrm{H}=\frac{2 \mathrm{G}^2}{2 \mathrm{~A}}\) \(\mathrm{H}=\frac{\mathrm{G}^2}{\mathrm{~A}}\)