2, 4, 8, 16, . . . a₂ - a₁ = 4 - 2 = 2 a₃ - a₂ 8 - 4 = 4 a₄ - a₃ = 16 - 8 = 8 The difference between the successive terms are not same Hence, it is not an A.P.
2, 4, 8, 16, . . .
a₂ – a₁ = 4 – 2 = 2
a₃ – a₂ 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
The difference between the successive terms are not same Hence, it is not an A.P.
a = 10, d = 10 First term a₁ = a = 10 Second term a₂ = a₁ + d = 10 + 10 = 20 Third term a₃ = a₂ + d = 20 + 10 = 30 Fourth term a₄ = a₃ + d = 30 + 10 = 40
a = 10, d = 10
First term a₁ = a = 10
Second term a₂ = a₁ + d = 10 + 10 = 20
Third term a₃ = a₂ + d = 20 + 10 = 30
Fourth term a₄ = a₃ + d = 30 + 10 = 40
3x² – 4√3x + 4 = 0 The given equations is of the form ax² + bx + c = 0 in which a = 3, b = -4√3, c = 4 Therefore, D = b² - 4ac = (-4√3)² - 4 × 3 × 4 = 48 - 48 = 0 So, the roots of quadratic equation are real and equal. Hence, x = 4√3±√0/6 = 4√3/6 = 2√3/3 [As x = -b±√b²-4ac] Hence, the roots of the qRead more
3x² – 4√3x + 4 = 0
The given equations is of the form ax² + bx + c = 0 in which a = 3, b = -4√3, c = 4
Therefore, D = b² – 4ac = (-4√3)² – 4 × 3 × 4 = 48 – 48 = 0
So, the roots of quadratic equation are real and equal.
Hence, x = 4√3±√0/6 = 4√3/6 = 2√3/3 [As x = -b±√b²-4ac]
Hence, the roots of the quadratic equation are 2√3/3 and 2√3/3.
kx (x - 2) + 6 = 0 On simplification, we get kx² - 2kx + 6 = 0 For the quadratic equation kx² - 2kx + 6 = 0 we have a = k, b = - 2k, c = 6. Therefore, b² - 4ac = (-2k)² - 4 × k × 6 = 4k² - 24k For equal roots, we have 4k² - 24k = 0 ⇒ 4k (k - 6) = 0 ⇒ 4k = 0 or (k - 6) = 0 ⇒ k = 0 or k = 6 But k ≠ 0,Read more
kx (x – 2) + 6 = 0
On simplification, we get kx² – 2kx + 6 = 0
For the quadratic equation kx² – 2kx + 6 = 0 we have a = k, b = – 2k, c = 6.
Therefore,
b² – 4ac = (-2k)² – 4 × k × 6 = 4k² – 24k
For equal roots, we have 4k² – 24k = 0
⇒ 4k (k – 6) = 0
⇒ 4k = 0 or (k – 6) = 0
⇒ k = 0 or k = 6
But k ≠ 0, as it doesn’t satisfies the equation kx(x -2) + 6 = 0.
Hence, k = 6
Which of the following are APs? If they form an AP, find the common difference d and write three more terms. 2, 4, 8, 16, . . .
2, 4, 8, 16, . . . a₂ - a₁ = 4 - 2 = 2 a₃ - a₂ 8 - 4 = 4 a₄ - a₃ = 16 - 8 = 8 The difference between the successive terms are not same Hence, it is not an A.P.
2, 4, 8, 16, . . .
See lessa₂ – a₁ = 4 – 2 = 2
a₃ – a₂ 8 – 4 = 4
a₄ – a₃ = 16 – 8 = 8
The difference between the successive terms are not same Hence, it is not an A.P.
For the following APs, write the first term and the common difference: 3, 1, – 1, – 3, . . .
First term a = 3 Common difference d = a₂-a₁ = 1 - 3 = -2
First term a = 3
See lessCommon difference d = a₂-a₁ = 1 – 3 = -2
Write first four terms of the AP, when the first term a and the common difference d are given as follows: a = 10, d = 10
a = 10, d = 10 First term a₁ = a = 10 Second term a₂ = a₁ + d = 10 + 10 = 20 Third term a₃ = a₂ + d = 20 + 10 = 30 Fourth term a₄ = a₃ + d = 30 + 10 = 40
a = 10, d = 10
See lessFirst term a₁ = a = 10
Second term a₂ = a₁ + d = 10 + 10 = 20
Third term a₃ = a₂ + d = 20 + 10 = 30
Fourth term a₄ = a₃ + d = 30 + 10 = 40
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: 3x² – 4√3 x + 4 = 0
3x² – 4√3x + 4 = 0 The given equations is of the form ax² + bx + c = 0 in which a = 3, b = -4√3, c = 4 Therefore, D = b² - 4ac = (-4√3)² - 4 × 3 × 4 = 48 - 48 = 0 So, the roots of quadratic equation are real and equal. Hence, x = 4√3±√0/6 = 4√3/6 = 2√3/3 [As x = -b±√b²-4ac] Hence, the roots of the qRead more
3x² – 4√3x + 4 = 0
See lessThe given equations is of the form ax² + bx + c = 0 in which a = 3, b = -4√3, c = 4
Therefore, D = b² – 4ac = (-4√3)² – 4 × 3 × 4 = 48 – 48 = 0
So, the roots of quadratic equation are real and equal.
Hence, x = 4√3±√0/6 = 4√3/6 = 2√3/3 [As x = -b±√b²-4ac]
Hence, the roots of the quadratic equation are 2√3/3 and 2√3/3.
Find the values of k for each of the following quadratic equations, so that they have two equal roots. kx (x – 2) + 6 = 0
kx (x - 2) + 6 = 0 On simplification, we get kx² - 2kx + 6 = 0 For the quadratic equation kx² - 2kx + 6 = 0 we have a = k, b = - 2k, c = 6. Therefore, b² - 4ac = (-2k)² - 4 × k × 6 = 4k² - 24k For equal roots, we have 4k² - 24k = 0 ⇒ 4k (k - 6) = 0 ⇒ 4k = 0 or (k - 6) = 0 ⇒ k = 0 or k = 6 But k ≠ 0,Read more
kx (x – 2) + 6 = 0
See lessOn simplification, we get kx² – 2kx + 6 = 0
For the quadratic equation kx² – 2kx + 6 = 0 we have a = k, b = – 2k, c = 6.
Therefore,
b² – 4ac = (-2k)² – 4 × k × 6 = 4k² – 24k
For equal roots, we have 4k² – 24k = 0
⇒ 4k (k – 6) = 0
⇒ 4k = 0 or (k – 6) = 0
⇒ k = 0 or k = 6
But k ≠ 0, as it doesn’t satisfies the equation kx(x -2) + 6 = 0.
Hence, k = 6