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Ex 5.3, 3 (iv) - Chapter 5 Class 10 Arithmetic Progressions
Last updated at May 29, 2023 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 You are here
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 (iv) Given a3 = 15, S10 = 125, find d and a10. For a3 = 15 We know that an = a + (n β 1) d Putting an = a3 = 15 & n = 3 15 = a + 2d 15 β 2d = a a = 15 β 2d For S10 = 125 We know that Sn = π/2(π+(πβ1)π) Putting Sn = S10 = 125, n = 10 S10 = 10/2 (2π+(10β1)π) 125 = 5 (2a + 9d) 125/5=2π+9π 25 = 2a + 9d 25 β 9d = 2a a = (ππ β ππ )/π From (1) and (2) 15 β 2d = (ππβππ )/π 30 β 4d = 25 β 9d 5 = β 5d 5/(β5)=π d = β1 Putting value of d in (1) a = 15 β 2d a = 15 β 2 Γ(β1) a = 15 + 2 a = 17 Now we need to find a10 an = a + (n β 1) d Putting n = 10 , a = 17 & d = β1 a10 = 17 + (10 β 1) Γ (β1) a10 = 17 + 9 Γ (β1) a10 = 17 β 9 a10 = 8
