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NCERT Solution Exercise 5.3 Class 10 Maths

Important formula for this Exercise: Sum of n terms of AP= Sn  (i) Sn = n/2[2a+(n-1)d] (ii) Sn = n/2[a + an ] (iii) Sn = n/2[a + l ] (iv) Sum of first n positive integers = Sn = n(n+1)/2 Q.1 Find the sum of the Following APs; (i) 2, 7,12,…………to 10 terms a = first term = 2…

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Mathematics

NCERT Solution Exercise 5.2 Class 10 Maths

Q1 Fill in the blanks in the following table, given that a is the first term d, the common difference and an the nth term of the AP: (i) a = 7 d = 3 n = 8 an = ? Solution: nth term of AP an = a +  (n-1)d a8 = 7 +…

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Mathematics

NCERT Solution Exercise 5.1 Class 10 Maths

Basic Concepts Sequence: Some numbers arranged in a definite order, according to a definite rule, are said to form a sequence. eg. (i) 1,2,3,…………. (ii) 100, 60, 30, …………… (iii) -1.0, -1.5, -2.0, -2.5, ………. Arithmetic Progressions(AP):  An arithmetic progression is a list of numbers on which each term is obtained by adding a fixed…

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Class 10th Coordinate Geometry

NCERT Solution Exercise 7.2 Class 10

Q.1 Find the coordinates of the point which divides the join of (-1,7) and (4,-3) in the ratio 2:3. Solution: Let A(-1,7) and B(4, -3) and Ratio m1 : m2 = 2 : 3 fig By Section Formula Coordinate of P(x, y) x = (m1x2 + m2x1)/m1+m2 ; y = (m1y2 + m2y1)/m1+m2 ⇒ x…

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Class 10th Coordinate Geometry | Mathematics

NCERT Solution Exercise 7.1 class 10

Q1. Find the distance between the Following pairs of points: (i) (2,3),(4,1)  Solution: Let By Distance Formula A B = √(x2 – x1)² + (y2 – y1)² ⇒ A B = √(4 – 2)² + (1 – 3)² ⇒ A B = √(2)² + (-2)² ⇒ A B = √4 + 4 ⇒ A B…

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Class 10 Quadratic Equations

NCERT Solution Exercise 4.4 Class10

Q1.Find the roots of the Following quadratic equations, if they exist, by the method of quadratic formula. Quadratic Formula/Shreedharacharya Sutra:- X =[-b ± √(b² – 4ac)]/2a Nature of Roots: Discriminant(D)= b² – 4ac 1. b²- 4ac < 0 Nature of Roots : Imaginary/No real root 2. b² – 4ac = 0 Nature of Roots: Real…

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Class 10 Quadratic Equations

NCERT Solution Exercise 4.2 Class10

Note: All questions Explained By Middle term splitting method/factorisation method. Q1. Find the roots if the Following quadratic equations by Factorisation:  (i) x² – 3x -10 = 0 Solution: x² – 3x -10 = 0 ⇒ x² – (5 – 2 )x – 10 = 0 ⇒ x² – 5x + 2x – 10 =…

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Class 10 Quadratic Equations

NCERT Solution Exercise 4.1 Class10

Q.1 Check whether the Following are quadratic equations: (i) (x+1)² = 2(x-3) x² + 2x + 1 = 2x – 6 ⇒ x² + 2x + 1 – 2x + 6 = 0 ⇒ x² + 7 = 0 by comparing a x² + b x + c = 0 a = 1, b =…

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NCERT Trigonometry Class 10 Exercise 8.3 Solutions

Trigonometric Ratios of Complementary Angles  Important identities: 1. sin(90 – θ) = cosθ 2. cos(90-θ) = sinθ 3. tan(90-θ) = cotθ 4. cot(90-θ) = secθ 5. sec(90-θ) = cosecθ 6. cosec(90-θ) = secθ Q.1 Evaluate the following: (i) sin18°/cos72° Solution: sin18°/cos72° = sin( 90°- 72° )/cos72°   [ sin(90-θ) = cosθ ] = cos 72° /…

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NCERT Solution Exercise 13.3 Class 10

Q.1 A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. Solution: Volume of cylinder = Volume of metallic sphere ⇒ πr²H       = 4/3 × πr³ ⇒ 6²H         = 4/3 × (4.2)³ ⇒…

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