Ex 5.2
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
Last updated at April 16, 2024 by Teachoo
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