Ex 5.3, 3 (ii) - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 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) You are here
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
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
Ex 5.3, 16 Important
Ex 5.3, 17
Ex 5.3, 18 Important
Ex 5.3, 19 Important
Ex 5.3, 20 Important
Chapter 5 Class 10 Arithmetic Progressions
Serial order wise
Transcript
Ex 5.3, 3 In an AP (ii) Given a = 7, a13 = 35, find d and S13. Given a = 7, a13 = 35 We need to find d We know that an = a + (n β 1) d Putting a = 7, n = 13 and an = 35 35 = 7 + (13 β 1) Γ π 35 = 7 + 12d 35 β 7 = 12d 28 = 12 d 28/12=π 7/3=π d = π/π Now we need to find S13 We can use formula Sn = π/π (π+π) Putting n = 13, a = 7, π = a13 = 35 = 13/2(7+35) = 13/2 Γ 42 = 13 Γ 21 = 273
