Ex 5.3, 7 - 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
Ex 5.3, 7 You are here
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, 7 Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. Given 22nd term = a22 = 149 Common difference = d = 7 We know that an = a + (n – 1) d Putting n = 22 , d = 7 , an = 149 149 = a + (22 – 1) × 7 149 = a + 21 × 7 149 = a + 147 149 = a + 147 149 – 147 = a 2 = a a = 2 Now, we need to find sum of first 22 terms And we are given a22 = 149 So, last term = l = 149 We use the formula Sn = 𝒏/𝟐 [ a + l ] Putting a = 2, n = 22, l = 149 = 22/2 [2 + 149] = 11 ×151 = 1661 Hence, Sum of first 22 terms is 1661
