Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 5.2, 20 - Chapter 5 Class 10 Arithmetic Progressions
Last updated at May 29, 2023 by Teachoo
Ex 5.2
Ex 5.2, 1
Ex 5.2, 2 (i) (MCQ)
Ex 5.2, 2 (ii) (MCQ) Important
Ex 5.2, 3 (i)
Ex 5.2, 3 (ii)
Ex 5.2, 3 (iii) Important
Ex 5.2, 3 (iv)
Ex 5.2, 3 (v) Important
Ex 5.2, 4
Ex 5.2, 5 (i)
Ex 5.2, 5 (ii) Important
Ex 5.2, 6 Important
Ex 5.2, 7
Ex 5.2, 8
Ex 5.2, 9
Ex 5.2, 10
Ex 5.2, 11
Ex 5.2, 12 Important
Ex 5.2, 13 Important
Ex 5.2, 14 Important
Ex 5.2, 15
Ex 5.2, 16
Ex 5.2, 17 Important
Ex 5.2, 18
Ex 5.2, 19
Ex 5.2, 20 Important You are here
Chapter 5 Class 10 Arithmetic Progressions
Serial order wise
Transcript
Ex 5.2, 20 Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n. Saving made first week = Rs 5 Saving made in second week = Rs 5 + 1.75 = Rs 6.75 Saving made in third week = 6. 75 + 1.75 = Rs 8.50 So, the series is 5, 6.75, 8.50 …… Since difference is same, it is an AP First Term = a = 5 Common difference = d = 1.75 Given, In nth week, her savings become Rs 20.75 So, an = 20.75 We need to find n We know that an = a + (n – 1) d Putting a = 5 , d = 1.75, an = 20.75 in formula 20.75 = 5 + (n – 1) × 1.75 20.75 – 5 = (n – 1) × 1.75 15.75 = (n – 1) × 1.75 15.75/1.75 = n – 1 1575/175 = n – 1 9 = n – 1 n – 1 = 9 n = 9 + 1 n = 10 Hence, in 10th week, her weekly savings become Rs 20.75
