118668
The first term of an A.P. is 148 and the common difference is \(\mathbf{- 2}\). If the A.M. of first \(n\) terms of the A.P. is 125 , then the value of \(n\) is
118669
If A.M. and G.M. of the roots of a quadratic equation are 8 and 5 respectively, then the quadratic equation is
1 \(\mathrm{x}^2+8 \mathrm{x}+5=0\)
2 \(x^2-16 x+10=0\)
3 \(x^2-16 x+25=0\)
4 \(x^2+8 x+25=0\)
5 \(x^2+10 x+15=0\)
Explanation:
C Let, \(\alpha\) and \(\beta\) be the two roots of the quadratic equation. Then, \(\frac{\alpha+\beta}{2}=8\) \(\alpha+\beta=16\) And, \(\sqrt{\alpha \beta}=5\) \(\alpha \beta=25\) Therefore, the required equation is: \(x^2-(\alpha+\beta) x+\alpha \beta=0\) On putting the value \((\alpha+\beta)\) and \(\alpha \beta\) in above equation, \(\mathrm{x}^2-16 \mathrm{x}+25=0\)
Kerala CEE-2019
Sequence and Series
118670
The arithmetic mean of \({ }^n C_0,{ }^n C_1,{ }^n C_2 \ldots . .{ }^n C_n\) is
118671
The arithmetic mean of two numbers \(x\) and \(y\) is 3 and geometric mean is 1 . Then \(x^2+y^2\) is equal to
1 30
2 31
3 32
4 33
5 34
Explanation:
E Given, Arithmetic mean of \(x, y=3\) \(\frac{x+y}{2}=3\) \(x+y=6\) Geometric mean of \(x, y=1\) \(\sqrt{\mathrm{xy}}=1\) \(\mathrm{xy}=1\) Now, \((x+y)^2=x^2+y^2+2 x y\) \(x^2+y^2=(x+y)^2-2 x y\) \(x^2+y^2=(6)^2-(2 \times 1)\) \(x^2+y^2=34\)
118668
The first term of an A.P. is 148 and the common difference is \(\mathbf{- 2}\). If the A.M. of first \(n\) terms of the A.P. is 125 , then the value of \(n\) is
118669
If A.M. and G.M. of the roots of a quadratic equation are 8 and 5 respectively, then the quadratic equation is
1 \(\mathrm{x}^2+8 \mathrm{x}+5=0\)
2 \(x^2-16 x+10=0\)
3 \(x^2-16 x+25=0\)
4 \(x^2+8 x+25=0\)
5 \(x^2+10 x+15=0\)
Explanation:
C Let, \(\alpha\) and \(\beta\) be the two roots of the quadratic equation. Then, \(\frac{\alpha+\beta}{2}=8\) \(\alpha+\beta=16\) And, \(\sqrt{\alpha \beta}=5\) \(\alpha \beta=25\) Therefore, the required equation is: \(x^2-(\alpha+\beta) x+\alpha \beta=0\) On putting the value \((\alpha+\beta)\) and \(\alpha \beta\) in above equation, \(\mathrm{x}^2-16 \mathrm{x}+25=0\)
Kerala CEE-2019
Sequence and Series
118670
The arithmetic mean of \({ }^n C_0,{ }^n C_1,{ }^n C_2 \ldots . .{ }^n C_n\) is
118671
The arithmetic mean of two numbers \(x\) and \(y\) is 3 and geometric mean is 1 . Then \(x^2+y^2\) is equal to
1 30
2 31
3 32
4 33
5 34
Explanation:
E Given, Arithmetic mean of \(x, y=3\) \(\frac{x+y}{2}=3\) \(x+y=6\) Geometric mean of \(x, y=1\) \(\sqrt{\mathrm{xy}}=1\) \(\mathrm{xy}=1\) Now, \((x+y)^2=x^2+y^2+2 x y\) \(x^2+y^2=(x+y)^2-2 x y\) \(x^2+y^2=(6)^2-(2 \times 1)\) \(x^2+y^2=34\)
118668
The first term of an A.P. is 148 and the common difference is \(\mathbf{- 2}\). If the A.M. of first \(n\) terms of the A.P. is 125 , then the value of \(n\) is
118669
If A.M. and G.M. of the roots of a quadratic equation are 8 and 5 respectively, then the quadratic equation is
1 \(\mathrm{x}^2+8 \mathrm{x}+5=0\)
2 \(x^2-16 x+10=0\)
3 \(x^2-16 x+25=0\)
4 \(x^2+8 x+25=0\)
5 \(x^2+10 x+15=0\)
Explanation:
C Let, \(\alpha\) and \(\beta\) be the two roots of the quadratic equation. Then, \(\frac{\alpha+\beta}{2}=8\) \(\alpha+\beta=16\) And, \(\sqrt{\alpha \beta}=5\) \(\alpha \beta=25\) Therefore, the required equation is: \(x^2-(\alpha+\beta) x+\alpha \beta=0\) On putting the value \((\alpha+\beta)\) and \(\alpha \beta\) in above equation, \(\mathrm{x}^2-16 \mathrm{x}+25=0\)
Kerala CEE-2019
Sequence and Series
118670
The arithmetic mean of \({ }^n C_0,{ }^n C_1,{ }^n C_2 \ldots . .{ }^n C_n\) is
118671
The arithmetic mean of two numbers \(x\) and \(y\) is 3 and geometric mean is 1 . Then \(x^2+y^2\) is equal to
1 30
2 31
3 32
4 33
5 34
Explanation:
E Given, Arithmetic mean of \(x, y=3\) \(\frac{x+y}{2}=3\) \(x+y=6\) Geometric mean of \(x, y=1\) \(\sqrt{\mathrm{xy}}=1\) \(\mathrm{xy}=1\) Now, \((x+y)^2=x^2+y^2+2 x y\) \(x^2+y^2=(x+y)^2-2 x y\) \(x^2+y^2=(6)^2-(2 \times 1)\) \(x^2+y^2=34\)
118668
The first term of an A.P. is 148 and the common difference is \(\mathbf{- 2}\). If the A.M. of first \(n\) terms of the A.P. is 125 , then the value of \(n\) is
118669
If A.M. and G.M. of the roots of a quadratic equation are 8 and 5 respectively, then the quadratic equation is
1 \(\mathrm{x}^2+8 \mathrm{x}+5=0\)
2 \(x^2-16 x+10=0\)
3 \(x^2-16 x+25=0\)
4 \(x^2+8 x+25=0\)
5 \(x^2+10 x+15=0\)
Explanation:
C Let, \(\alpha\) and \(\beta\) be the two roots of the quadratic equation. Then, \(\frac{\alpha+\beta}{2}=8\) \(\alpha+\beta=16\) And, \(\sqrt{\alpha \beta}=5\) \(\alpha \beta=25\) Therefore, the required equation is: \(x^2-(\alpha+\beta) x+\alpha \beta=0\) On putting the value \((\alpha+\beta)\) and \(\alpha \beta\) in above equation, \(\mathrm{x}^2-16 \mathrm{x}+25=0\)
Kerala CEE-2019
Sequence and Series
118670
The arithmetic mean of \({ }^n C_0,{ }^n C_1,{ }^n C_2 \ldots . .{ }^n C_n\) is
118671
The arithmetic mean of two numbers \(x\) and \(y\) is 3 and geometric mean is 1 . Then \(x^2+y^2\) is equal to
1 30
2 31
3 32
4 33
5 34
Explanation:
E Given, Arithmetic mean of \(x, y=3\) \(\frac{x+y}{2}=3\) \(x+y=6\) Geometric mean of \(x, y=1\) \(\sqrt{\mathrm{xy}}=1\) \(\mathrm{xy}=1\) Now, \((x+y)^2=x^2+y^2+2 x y\) \(x^2+y^2=(x+y)^2-2 x y\) \(x^2+y^2=(6)^2-(2 \times 1)\) \(x^2+y^2=34\)
