118683
If \(a>0, b>0, c>0\) and \(a, b, c\), are distinct, then \((a+b)(b+c)(c+a)\) is greater than
1 \(2(a+b+c)\)
2 \(3(a+b+c)\)
3 \(6 \mathrm{abc}\)
4 \(8 \mathrm{abc}\)
Explanation:
D Let, the numbers are \(\mathrm{a}\) and \(\mathrm{b}\) \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) \(\frac{\mathrm{b}+\mathrm{c}}{2} \geq \sqrt{\mathrm{bc}}\) For a, c \(\frac{\mathrm{a}+\mathrm{c}}{2} \geq \sqrt{\mathrm{ac}}\) On multiplying equation (i), (ii) and (iii) we get - \(\frac{(a+b)}{2} \cdot \frac{(b+c)}{2} \cdot \frac{(c+a)}{2} \geq \sqrt{a b} \sqrt{b c} \sqrt{a c}\) \(\frac{(a+b)(b+c)(c+a)}{8} \geq \sqrt{a^2 b^2 c^2}\) \((a+b)(b+c)(c+a) \geq 8 a b c\)
BITSAT-2020
Sequence and Series
118684
If \(a_1, a_2\), and \(a_3\) be any positive real numbers, then which of the following statement is true?
D We know that, Geometric mean, \((\mathrm{GM})=\sqrt[n]{\mathrm{x}_1 \cdot \mathrm{x}_2 \mathrm{x}_3 \ldots . \mathrm{x}_{\mathrm{n}}}\) Harmonic mean \((\mathrm{HM})=\frac{\mathrm{n}}{\left(\frac{1}{\mathrm{x}_1}+\frac{1}{\mathrm{x}_2}+\ldots .+\frac{1}{\mathrm{x}_{\mathrm{n}}}\right)}\) \(a_1, a_2\) and \(a_3\) are positive real numbers. So, \(\mathrm{GM}=\sqrt[3]{\mathrm{a}_1 \times \mathrm{a}_2 \times \mathrm{a}_3}\) \(\mathrm{HM}=\frac{3}{\left(\frac{1}{\mathrm{a}_1}+\frac{1}{\mathrm{a}_2}+\frac{1}{\mathrm{a}_3}\right)}\) Since, \(\mathrm{GM} \geq \mathrm{HM}\) \(\therefore \sqrt[3]{a_1 \times a_2 \times a_3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 \times a_2 \times a_3\right)^{1 / 3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 a_2 a_3\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)^3 \geq 27\)
VITEEE-2013
Sequence and Series
118685
The harmonic mean of two numbers is 4 and the arithmetic and geometric mean satisfy the relation \(2 \mathbf{A}+\mathbf{G}^{\mathbf{2}}=\mathbf{2 7}\), the numbers are
1 6,3
2 5,4
3 \(5,-2.5\)
4 \(-3,1\)
Explanation:
A Given, \(2 \mathrm{~A}+\mathrm{G}^2=27\) Let, the numbers be \(a\) and \(b\). According to question harmonic mean of these numbers- \(\frac{2}{\frac{1}{a}+\frac{1}{b}}=4\) \(\frac{2 a b}{a+b}=4\) \(2 a b=4(a+b)\) Now, arithmetic and geometric mean of \(a\) and \(b\) \(\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2} \text { and } \mathrm{G}=\sqrt{\mathrm{ab}}\) On putting the value of \(\mathrm{A}\) and \(\mathrm{G}\) in equation (i) \(2\left(\frac{a+b}{2}\right)+(\sqrt{a b})^2=27\) \((a+b)+a b=27\) \((a+b)=27-a b\) On putting the value of \((a+b)\) in equation (ii) we get - \(2 \mathrm{ab}=4(27-\mathrm{ab})\) \(\mathrm{ab}=54-2 \mathrm{ab}\) \(3 \mathrm{ab}=54\) \(\mathrm{ab}=18\) On putting the value of \(a b\) in equation (iii) we get - \(a+b=27-18=9\) \(a+b=9\) And, \(\quad \mathrm{ab}=18\) On solving equation (iv) and (v) we get - \(\mathrm{a}=6 \text { and } \mathrm{b}=3\)
UPSEE-2012
Sequence and Series
118686
Let \(a_1, a_2, . ., a_{10}\), be in AP and \(h_1, h_2, \ldots ., h_{10}\) be in HP. If \(a_1=h_1=2\) and \(a_{10}=h_{10}=3\). Then, \(a_4 h_7\) is
1 2
2 3
3 5
4 6
Explanation:
D Given, \(\mathrm{a}_1=\mathrm{h}_1=2\) and \(\mathrm{a}_{10}=\mathrm{h}_{10}=3\) We know that, \(n^{\text {th }}\) term of A.P. \(T_n=a+(n-1) d\) \(a_{10}=a_1+(10-1) d\) \(\left(a_{10}-a_1\right)=9 d\) \((3-2)=9 d\) \(d=\left(\frac{3-2}{9}\right)=\frac{1}{9}\) So, \(\mathrm{a}_4=\mathrm{a}_1+(4-1) \mathrm{d}\) \(\mathrm{a}_4=2+3 \times \frac{1}{9}\) \(\mathrm{a}_4=2+\frac{1}{3}\) \(\mathrm{a}_4=\frac{7}{3}\) We know that \(\mathrm{n}^{\text {th }}\) term of H.P. \(\left(\frac{1}{\mathrm{~h}_{10}}\right)=\left(\frac{1}{\mathrm{~h}_1}\right)+(\mathrm{n}-1) \mathrm{d}_1\) \(\frac{1}{3}=\frac{1}{2}+9 \mathrm{~d}_1\) \(\frac{-1}{6}=9 \mathrm{~d}_1\) \(\mathrm{~d}_1=-\frac{1}{54}\) Now, \(\mathrm{h}_7\) of H.P. \(\frac{1}{\mathrm{~h}_7}=\frac{1}{\mathrm{~h}_1}+(7-1) \mathrm{d}_1\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}+6 \mathrm{~d}_1=\frac{1}{2}+6 \times\left(\frac{-1}{54}\right)\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}-\frac{1}{9}\) \(\frac{1}{\mathrm{~h}_7}=\frac{7}{18}\) \(\mathrm{~h}_7=\frac{18}{7}\) According to equation \(\mathrm{a}_4 \times \mathrm{h}_7=\frac{7}{3} \times \frac{18}{7}\) \(\mathrm{a}_4 \mathrm{~h}_7=6\)
118683
If \(a>0, b>0, c>0\) and \(a, b, c\), are distinct, then \((a+b)(b+c)(c+a)\) is greater than
1 \(2(a+b+c)\)
2 \(3(a+b+c)\)
3 \(6 \mathrm{abc}\)
4 \(8 \mathrm{abc}\)
Explanation:
D Let, the numbers are \(\mathrm{a}\) and \(\mathrm{b}\) \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) \(\frac{\mathrm{b}+\mathrm{c}}{2} \geq \sqrt{\mathrm{bc}}\) For a, c \(\frac{\mathrm{a}+\mathrm{c}}{2} \geq \sqrt{\mathrm{ac}}\) On multiplying equation (i), (ii) and (iii) we get - \(\frac{(a+b)}{2} \cdot \frac{(b+c)}{2} \cdot \frac{(c+a)}{2} \geq \sqrt{a b} \sqrt{b c} \sqrt{a c}\) \(\frac{(a+b)(b+c)(c+a)}{8} \geq \sqrt{a^2 b^2 c^2}\) \((a+b)(b+c)(c+a) \geq 8 a b c\)
BITSAT-2020
Sequence and Series
118684
If \(a_1, a_2\), and \(a_3\) be any positive real numbers, then which of the following statement is true?
D We know that, Geometric mean, \((\mathrm{GM})=\sqrt[n]{\mathrm{x}_1 \cdot \mathrm{x}_2 \mathrm{x}_3 \ldots . \mathrm{x}_{\mathrm{n}}}\) Harmonic mean \((\mathrm{HM})=\frac{\mathrm{n}}{\left(\frac{1}{\mathrm{x}_1}+\frac{1}{\mathrm{x}_2}+\ldots .+\frac{1}{\mathrm{x}_{\mathrm{n}}}\right)}\) \(a_1, a_2\) and \(a_3\) are positive real numbers. So, \(\mathrm{GM}=\sqrt[3]{\mathrm{a}_1 \times \mathrm{a}_2 \times \mathrm{a}_3}\) \(\mathrm{HM}=\frac{3}{\left(\frac{1}{\mathrm{a}_1}+\frac{1}{\mathrm{a}_2}+\frac{1}{\mathrm{a}_3}\right)}\) Since, \(\mathrm{GM} \geq \mathrm{HM}\) \(\therefore \sqrt[3]{a_1 \times a_2 \times a_3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 \times a_2 \times a_3\right)^{1 / 3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 a_2 a_3\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)^3 \geq 27\)
VITEEE-2013
Sequence and Series
118685
The harmonic mean of two numbers is 4 and the arithmetic and geometric mean satisfy the relation \(2 \mathbf{A}+\mathbf{G}^{\mathbf{2}}=\mathbf{2 7}\), the numbers are
1 6,3
2 5,4
3 \(5,-2.5\)
4 \(-3,1\)
Explanation:
A Given, \(2 \mathrm{~A}+\mathrm{G}^2=27\) Let, the numbers be \(a\) and \(b\). According to question harmonic mean of these numbers- \(\frac{2}{\frac{1}{a}+\frac{1}{b}}=4\) \(\frac{2 a b}{a+b}=4\) \(2 a b=4(a+b)\) Now, arithmetic and geometric mean of \(a\) and \(b\) \(\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2} \text { and } \mathrm{G}=\sqrt{\mathrm{ab}}\) On putting the value of \(\mathrm{A}\) and \(\mathrm{G}\) in equation (i) \(2\left(\frac{a+b}{2}\right)+(\sqrt{a b})^2=27\) \((a+b)+a b=27\) \((a+b)=27-a b\) On putting the value of \((a+b)\) in equation (ii) we get - \(2 \mathrm{ab}=4(27-\mathrm{ab})\) \(\mathrm{ab}=54-2 \mathrm{ab}\) \(3 \mathrm{ab}=54\) \(\mathrm{ab}=18\) On putting the value of \(a b\) in equation (iii) we get - \(a+b=27-18=9\) \(a+b=9\) And, \(\quad \mathrm{ab}=18\) On solving equation (iv) and (v) we get - \(\mathrm{a}=6 \text { and } \mathrm{b}=3\)
UPSEE-2012
Sequence and Series
118686
Let \(a_1, a_2, . ., a_{10}\), be in AP and \(h_1, h_2, \ldots ., h_{10}\) be in HP. If \(a_1=h_1=2\) and \(a_{10}=h_{10}=3\). Then, \(a_4 h_7\) is
1 2
2 3
3 5
4 6
Explanation:
D Given, \(\mathrm{a}_1=\mathrm{h}_1=2\) and \(\mathrm{a}_{10}=\mathrm{h}_{10}=3\) We know that, \(n^{\text {th }}\) term of A.P. \(T_n=a+(n-1) d\) \(a_{10}=a_1+(10-1) d\) \(\left(a_{10}-a_1\right)=9 d\) \((3-2)=9 d\) \(d=\left(\frac{3-2}{9}\right)=\frac{1}{9}\) So, \(\mathrm{a}_4=\mathrm{a}_1+(4-1) \mathrm{d}\) \(\mathrm{a}_4=2+3 \times \frac{1}{9}\) \(\mathrm{a}_4=2+\frac{1}{3}\) \(\mathrm{a}_4=\frac{7}{3}\) We know that \(\mathrm{n}^{\text {th }}\) term of H.P. \(\left(\frac{1}{\mathrm{~h}_{10}}\right)=\left(\frac{1}{\mathrm{~h}_1}\right)+(\mathrm{n}-1) \mathrm{d}_1\) \(\frac{1}{3}=\frac{1}{2}+9 \mathrm{~d}_1\) \(\frac{-1}{6}=9 \mathrm{~d}_1\) \(\mathrm{~d}_1=-\frac{1}{54}\) Now, \(\mathrm{h}_7\) of H.P. \(\frac{1}{\mathrm{~h}_7}=\frac{1}{\mathrm{~h}_1}+(7-1) \mathrm{d}_1\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}+6 \mathrm{~d}_1=\frac{1}{2}+6 \times\left(\frac{-1}{54}\right)\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}-\frac{1}{9}\) \(\frac{1}{\mathrm{~h}_7}=\frac{7}{18}\) \(\mathrm{~h}_7=\frac{18}{7}\) According to equation \(\mathrm{a}_4 \times \mathrm{h}_7=\frac{7}{3} \times \frac{18}{7}\) \(\mathrm{a}_4 \mathrm{~h}_7=6\)
118683
If \(a>0, b>0, c>0\) and \(a, b, c\), are distinct, then \((a+b)(b+c)(c+a)\) is greater than
1 \(2(a+b+c)\)
2 \(3(a+b+c)\)
3 \(6 \mathrm{abc}\)
4 \(8 \mathrm{abc}\)
Explanation:
D Let, the numbers are \(\mathrm{a}\) and \(\mathrm{b}\) \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) \(\frac{\mathrm{b}+\mathrm{c}}{2} \geq \sqrt{\mathrm{bc}}\) For a, c \(\frac{\mathrm{a}+\mathrm{c}}{2} \geq \sqrt{\mathrm{ac}}\) On multiplying equation (i), (ii) and (iii) we get - \(\frac{(a+b)}{2} \cdot \frac{(b+c)}{2} \cdot \frac{(c+a)}{2} \geq \sqrt{a b} \sqrt{b c} \sqrt{a c}\) \(\frac{(a+b)(b+c)(c+a)}{8} \geq \sqrt{a^2 b^2 c^2}\) \((a+b)(b+c)(c+a) \geq 8 a b c\)
BITSAT-2020
Sequence and Series
118684
If \(a_1, a_2\), and \(a_3\) be any positive real numbers, then which of the following statement is true?
D We know that, Geometric mean, \((\mathrm{GM})=\sqrt[n]{\mathrm{x}_1 \cdot \mathrm{x}_2 \mathrm{x}_3 \ldots . \mathrm{x}_{\mathrm{n}}}\) Harmonic mean \((\mathrm{HM})=\frac{\mathrm{n}}{\left(\frac{1}{\mathrm{x}_1}+\frac{1}{\mathrm{x}_2}+\ldots .+\frac{1}{\mathrm{x}_{\mathrm{n}}}\right)}\) \(a_1, a_2\) and \(a_3\) are positive real numbers. So, \(\mathrm{GM}=\sqrt[3]{\mathrm{a}_1 \times \mathrm{a}_2 \times \mathrm{a}_3}\) \(\mathrm{HM}=\frac{3}{\left(\frac{1}{\mathrm{a}_1}+\frac{1}{\mathrm{a}_2}+\frac{1}{\mathrm{a}_3}\right)}\) Since, \(\mathrm{GM} \geq \mathrm{HM}\) \(\therefore \sqrt[3]{a_1 \times a_2 \times a_3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 \times a_2 \times a_3\right)^{1 / 3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 a_2 a_3\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)^3 \geq 27\)
VITEEE-2013
Sequence and Series
118685
The harmonic mean of two numbers is 4 and the arithmetic and geometric mean satisfy the relation \(2 \mathbf{A}+\mathbf{G}^{\mathbf{2}}=\mathbf{2 7}\), the numbers are
1 6,3
2 5,4
3 \(5,-2.5\)
4 \(-3,1\)
Explanation:
A Given, \(2 \mathrm{~A}+\mathrm{G}^2=27\) Let, the numbers be \(a\) and \(b\). According to question harmonic mean of these numbers- \(\frac{2}{\frac{1}{a}+\frac{1}{b}}=4\) \(\frac{2 a b}{a+b}=4\) \(2 a b=4(a+b)\) Now, arithmetic and geometric mean of \(a\) and \(b\) \(\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2} \text { and } \mathrm{G}=\sqrt{\mathrm{ab}}\) On putting the value of \(\mathrm{A}\) and \(\mathrm{G}\) in equation (i) \(2\left(\frac{a+b}{2}\right)+(\sqrt{a b})^2=27\) \((a+b)+a b=27\) \((a+b)=27-a b\) On putting the value of \((a+b)\) in equation (ii) we get - \(2 \mathrm{ab}=4(27-\mathrm{ab})\) \(\mathrm{ab}=54-2 \mathrm{ab}\) \(3 \mathrm{ab}=54\) \(\mathrm{ab}=18\) On putting the value of \(a b\) in equation (iii) we get - \(a+b=27-18=9\) \(a+b=9\) And, \(\quad \mathrm{ab}=18\) On solving equation (iv) and (v) we get - \(\mathrm{a}=6 \text { and } \mathrm{b}=3\)
UPSEE-2012
Sequence and Series
118686
Let \(a_1, a_2, . ., a_{10}\), be in AP and \(h_1, h_2, \ldots ., h_{10}\) be in HP. If \(a_1=h_1=2\) and \(a_{10}=h_{10}=3\). Then, \(a_4 h_7\) is
1 2
2 3
3 5
4 6
Explanation:
D Given, \(\mathrm{a}_1=\mathrm{h}_1=2\) and \(\mathrm{a}_{10}=\mathrm{h}_{10}=3\) We know that, \(n^{\text {th }}\) term of A.P. \(T_n=a+(n-1) d\) \(a_{10}=a_1+(10-1) d\) \(\left(a_{10}-a_1\right)=9 d\) \((3-2)=9 d\) \(d=\left(\frac{3-2}{9}\right)=\frac{1}{9}\) So, \(\mathrm{a}_4=\mathrm{a}_1+(4-1) \mathrm{d}\) \(\mathrm{a}_4=2+3 \times \frac{1}{9}\) \(\mathrm{a}_4=2+\frac{1}{3}\) \(\mathrm{a}_4=\frac{7}{3}\) We know that \(\mathrm{n}^{\text {th }}\) term of H.P. \(\left(\frac{1}{\mathrm{~h}_{10}}\right)=\left(\frac{1}{\mathrm{~h}_1}\right)+(\mathrm{n}-1) \mathrm{d}_1\) \(\frac{1}{3}=\frac{1}{2}+9 \mathrm{~d}_1\) \(\frac{-1}{6}=9 \mathrm{~d}_1\) \(\mathrm{~d}_1=-\frac{1}{54}\) Now, \(\mathrm{h}_7\) of H.P. \(\frac{1}{\mathrm{~h}_7}=\frac{1}{\mathrm{~h}_1}+(7-1) \mathrm{d}_1\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}+6 \mathrm{~d}_1=\frac{1}{2}+6 \times\left(\frac{-1}{54}\right)\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}-\frac{1}{9}\) \(\frac{1}{\mathrm{~h}_7}=\frac{7}{18}\) \(\mathrm{~h}_7=\frac{18}{7}\) According to equation \(\mathrm{a}_4 \times \mathrm{h}_7=\frac{7}{3} \times \frac{18}{7}\) \(\mathrm{a}_4 \mathrm{~h}_7=6\)
118683
If \(a>0, b>0, c>0\) and \(a, b, c\), are distinct, then \((a+b)(b+c)(c+a)\) is greater than
1 \(2(a+b+c)\)
2 \(3(a+b+c)\)
3 \(6 \mathrm{abc}\)
4 \(8 \mathrm{abc}\)
Explanation:
D Let, the numbers are \(\mathrm{a}\) and \(\mathrm{b}\) \(\mathrm{AM} \geq \mathrm{GM}\) \(\frac{\mathrm{a}+\mathrm{b}}{2} \geq \sqrt{\mathrm{ab}}\) \(\frac{\mathrm{b}+\mathrm{c}}{2} \geq \sqrt{\mathrm{bc}}\) For a, c \(\frac{\mathrm{a}+\mathrm{c}}{2} \geq \sqrt{\mathrm{ac}}\) On multiplying equation (i), (ii) and (iii) we get - \(\frac{(a+b)}{2} \cdot \frac{(b+c)}{2} \cdot \frac{(c+a)}{2} \geq \sqrt{a b} \sqrt{b c} \sqrt{a c}\) \(\frac{(a+b)(b+c)(c+a)}{8} \geq \sqrt{a^2 b^2 c^2}\) \((a+b)(b+c)(c+a) \geq 8 a b c\)
BITSAT-2020
Sequence and Series
118684
If \(a_1, a_2\), and \(a_3\) be any positive real numbers, then which of the following statement is true?
D We know that, Geometric mean, \((\mathrm{GM})=\sqrt[n]{\mathrm{x}_1 \cdot \mathrm{x}_2 \mathrm{x}_3 \ldots . \mathrm{x}_{\mathrm{n}}}\) Harmonic mean \((\mathrm{HM})=\frac{\mathrm{n}}{\left(\frac{1}{\mathrm{x}_1}+\frac{1}{\mathrm{x}_2}+\ldots .+\frac{1}{\mathrm{x}_{\mathrm{n}}}\right)}\) \(a_1, a_2\) and \(a_3\) are positive real numbers. So, \(\mathrm{GM}=\sqrt[3]{\mathrm{a}_1 \times \mathrm{a}_2 \times \mathrm{a}_3}\) \(\mathrm{HM}=\frac{3}{\left(\frac{1}{\mathrm{a}_1}+\frac{1}{\mathrm{a}_2}+\frac{1}{\mathrm{a}_3}\right)}\) Since, \(\mathrm{GM} \geq \mathrm{HM}\) \(\therefore \sqrt[3]{a_1 \times a_2 \times a_3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 \times a_2 \times a_3\right)^{1 / 3} \geq \frac{3}{\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)}\) \(\left(a_1 a_2 a_3\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}\right)^3 \geq 27\)
VITEEE-2013
Sequence and Series
118685
The harmonic mean of two numbers is 4 and the arithmetic and geometric mean satisfy the relation \(2 \mathbf{A}+\mathbf{G}^{\mathbf{2}}=\mathbf{2 7}\), the numbers are
1 6,3
2 5,4
3 \(5,-2.5\)
4 \(-3,1\)
Explanation:
A Given, \(2 \mathrm{~A}+\mathrm{G}^2=27\) Let, the numbers be \(a\) and \(b\). According to question harmonic mean of these numbers- \(\frac{2}{\frac{1}{a}+\frac{1}{b}}=4\) \(\frac{2 a b}{a+b}=4\) \(2 a b=4(a+b)\) Now, arithmetic and geometric mean of \(a\) and \(b\) \(\mathrm{A}=\frac{\mathrm{a}+\mathrm{b}}{2} \text { and } \mathrm{G}=\sqrt{\mathrm{ab}}\) On putting the value of \(\mathrm{A}\) and \(\mathrm{G}\) in equation (i) \(2\left(\frac{a+b}{2}\right)+(\sqrt{a b})^2=27\) \((a+b)+a b=27\) \((a+b)=27-a b\) On putting the value of \((a+b)\) in equation (ii) we get - \(2 \mathrm{ab}=4(27-\mathrm{ab})\) \(\mathrm{ab}=54-2 \mathrm{ab}\) \(3 \mathrm{ab}=54\) \(\mathrm{ab}=18\) On putting the value of \(a b\) in equation (iii) we get - \(a+b=27-18=9\) \(a+b=9\) And, \(\quad \mathrm{ab}=18\) On solving equation (iv) and (v) we get - \(\mathrm{a}=6 \text { and } \mathrm{b}=3\)
UPSEE-2012
Sequence and Series
118686
Let \(a_1, a_2, . ., a_{10}\), be in AP and \(h_1, h_2, \ldots ., h_{10}\) be in HP. If \(a_1=h_1=2\) and \(a_{10}=h_{10}=3\). Then, \(a_4 h_7\) is
1 2
2 3
3 5
4 6
Explanation:
D Given, \(\mathrm{a}_1=\mathrm{h}_1=2\) and \(\mathrm{a}_{10}=\mathrm{h}_{10}=3\) We know that, \(n^{\text {th }}\) term of A.P. \(T_n=a+(n-1) d\) \(a_{10}=a_1+(10-1) d\) \(\left(a_{10}-a_1\right)=9 d\) \((3-2)=9 d\) \(d=\left(\frac{3-2}{9}\right)=\frac{1}{9}\) So, \(\mathrm{a}_4=\mathrm{a}_1+(4-1) \mathrm{d}\) \(\mathrm{a}_4=2+3 \times \frac{1}{9}\) \(\mathrm{a}_4=2+\frac{1}{3}\) \(\mathrm{a}_4=\frac{7}{3}\) We know that \(\mathrm{n}^{\text {th }}\) term of H.P. \(\left(\frac{1}{\mathrm{~h}_{10}}\right)=\left(\frac{1}{\mathrm{~h}_1}\right)+(\mathrm{n}-1) \mathrm{d}_1\) \(\frac{1}{3}=\frac{1}{2}+9 \mathrm{~d}_1\) \(\frac{-1}{6}=9 \mathrm{~d}_1\) \(\mathrm{~d}_1=-\frac{1}{54}\) Now, \(\mathrm{h}_7\) of H.P. \(\frac{1}{\mathrm{~h}_7}=\frac{1}{\mathrm{~h}_1}+(7-1) \mathrm{d}_1\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}+6 \mathrm{~d}_1=\frac{1}{2}+6 \times\left(\frac{-1}{54}\right)\) \(\frac{1}{\mathrm{~h}_7}=\frac{1}{2}-\frac{1}{9}\) \(\frac{1}{\mathrm{~h}_7}=\frac{7}{18}\) \(\mathrm{~h}_7=\frac{18}{7}\) According to equation \(\mathrm{a}_4 \times \mathrm{h}_7=\frac{7}{3} \times \frac{18}{7}\) \(\mathrm{a}_4 \mathrm{~h}_7=6\)
