Houses of a row numbered consecutively 1 to 49. Show that there is

Ex 5.4, 4 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 2

Ex 5.4, 4 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 3

Ex 5.4, 4 (Optional) - Chapter 5 Class 10 Arithmetic Progressions - Part 4

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Ex 5.4, 4 (Optional) - 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 3 = S5 − S3 = S5 − SX Since, Sum of numbers of houses preceding x = Sum of the number of houses following x Sx −1 = S5 − Sx Ex 5.4, 4 (Optional) 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 Sum of first n natural numbers = Sn = (𝒏(𝒏 + 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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo