

Ex 5.3
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) You are here
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
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
Last updated at Aug. 3, 2021 by Teachoo
Ex 5.3, 3 In an AP (iii) Given a12 = 37, d = 3, find a and S12. Given a12 = 37, d = 3 Finding a We know that an = a + (n – 1) d Putting d = 3, n = 12 and a12 = 37 37 = a + (12 – 1) × 3 37 = a + 11 × 3 37 = a + 33 37 − 33 = a 4 = a a = 4 Now, we can find (S12) by using formula Sn = 𝒏/𝟐 (𝒂+𝒍) Putting n = 12 , a = 4, 𝑙 = a12 = 37 S12 = 12/2 (4+37) = 12/2×41 = 6 ×41 = 246