Ex 5.3, 3 (vii) - 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)
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) You are here
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 (vii) Given a = 8, an = 62, Sn = 210, find n and d. Given a = 8, an = 62, Sn = 210, Since there are n terms, π = an = 62 We use the formula Sn = π/π (π+π) Putting a = 8, Sn = 210, π = an = 62 210 = π/2 (8+62) 210 Γ 2=π Γ (70) 420 = n Γ 70 420/70=π 6 = n n = 6 Now we need to find d We can use formula an = a + (n β 1) d Putting an = 62, a = 8, n = 6 62 = 8 + (6 β 1) Γ π 62 = 8 + 5d 62 β 8 = 5d 54 = 5d d = ππ/π
