Ex 5.3, 6 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at March 22, 2021 by Teachoo
Ex 5.3
Ex 5.3, 1 (i)
Ex 5.3, 1 (ii)
Ex 5.3, 1 (iii) Important
Ex 5.3, 1 (iv)
Ex 5.3, 2 (i)
Ex 5.3, 2 (ii)
Ex 5.3, 2 (iii) Important
Ex 5.3, 3 (i)
Ex 5.3, 3 (ii)
Ex 5.3, 3 (iii)
Ex 5.3, 3 (iv) Important
Ex 5.3, 3 (v)
Ex 5.3, 3 (vi) Important
Ex 5.3, 3 (vii)
Ex 5.3, 3 (viii) Important
Ex 5.3, 3 (ix)
Ex 5.3, 3 (x)
Ex 5.3, 4
Ex 5.3, 5
Ex 5.3, 6 Important You are here
Ex 5.3, 7
Ex 5.3, 8
Ex 5.3, 9
Ex 5.3, 10 (i)
Ex 5.3, 10 (ii) Important
Ex 5.3, 11 Important
Ex 5.3, 12
Ex 5.3, 13
Ex 5.3, 14 Important
Ex 5.3, 15 Deleted for CBSE Board 2022 Exams
Ex 5.3, 16 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 17 Deleted for CBSE Board 2022 Exams
Ex 5.3, 18 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 19 Important Deleted for CBSE Board 2022 Exams
Ex 5.3, 20 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
Transcript
Ex 5.3, 6 The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Given a = 17 , l = 350, d = 9 We need to find how many terms i.e. n and their sum i.e. Sn We know that an = a + (n − 1)d Putting a = 17 ,an = l = 350, d = 9 350 = 17 + (n − 1) × 9 350 − 17 = (n − 1) × 9 333 = (n − 1) × 9 333/9 = (n − 1) 37 = n − 1 n − 1 = 37 n = 38 To find sum, We use the formula Sum = 𝒏/𝟐 (a + l) Putting n = 38, a = 17, l = 350 Sum = 38/2 (17 + 350) Sum = 19 × 367 Sum = 6973
