Learn all Concepts of Chapter 5 Class 10 (with VIDEOS). Check - Arithmetic Progressions - Class 10    1. Chapter 5 Class 10 Arithmetic Progressions
2. Serial order wise
3. Ex 5.4 (Optional)

Transcript

Ex 5.4, 4 (Introduction) The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint : Sx – 1 = S49 – Sx] Introduction: Let 5 houses numbered from 1 to 5 be arranged in a row 1 2 3 4 5 Now, let x = 3 Then Sum of number of houses preceding house numbered x = Sum of number of houses preceding house numbered 3 = S2 = S3 − 1 = Sx − 1 Sum of number of houses following x = Sum of number of houses following 2 = S5 − S3 = S5 − SX Since, Sum of numbers of houses preceding x = Sum of the number of houses following x Sx −1 = S5 − Sx Since, Sum of numbers of houses preceding x = Sum of the number of houses following x Sx−1 = S49 − Sx ((𝑥 − 1)(𝑥 − 1 + 1))/2 = (49 (49 + 1))/2 − (𝑥 (𝑥 + 1))/2 (𝑥(𝑥 − 1) )/2 = (49 × 50 )/2 − (𝑥 (𝑥 + 1))/2 Ex 5.4, 4 The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint : Sx – 1 = S49 – Sx] Since, Sum of numbers of houses preceding x = Sum of the number of houses following x Sx−1 = S49 − Sx ((𝑥 − 1)(𝑥 − 1 + 1))/2 = (49 (49 + 1))/2 − (𝑥 (𝑥 + 1))/2 (𝑥(𝑥 − 1) )/2 = (49 × 50 )/2 − (𝑥 (𝑥 + 1))/2 1/2 [𝑥(𝑥−1)] = 1/2 [2450−𝑥(𝑥+1)] x(x − 1) = 2450 − x(x + 1) x2 − x = 2450 − x2 − x x2 − x + x2 + x = 2450 2x2 = 2450 x2 = 1225 x = √1225 x = √(52" × 72" ) x = √352 x = ± 35 Since number of houses can’t be negative ∴ x = 35

Ex 5.4 (Optional) 