Ex 5.3

Chapter 5 Class 10 Arithmetic Progressions
Serial order wise

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Transcript

Ex 5.3, 3 In an AP (viii) Given an = 4, d = 2, Sn = β14, find n and a. Given an = 4, d = 2, Sn = β14 Since there are n terms, π = an = 4 We use the formula Sn = π/π (π+π) Putting Sn = β14, π = an = 4 β14 = π/2 (π+4) β14 Γ 2 =π(π+4) β28 = n (a + 4) (β28)/(π + 4)=π n = (βππ)/(π + π) Also we know that an = a + (n β 1) d Putting an = 4 , d = 2 4 = a + (n β 1) Γ 2 4 = a + 2n β 2 4 + 2 = a + 2n 6 = a + 2n Putting n = (β ππ)/(π + π) 6 = a + 2((β 28)/(π + 4)) 6 = a β 56/(π + 4) 6 = (π(π + 4) β 56)/(π + 4) 6 = (π2 + 4π β 56)/(π + 4) 6 (a + 4) = a2 + 4a β 56 6a + 24 = a2 4a β 56 0 = a2 + 4a β 56 β 6a β 24 0 = a2 β 2a β 80 a2 β 2a β 80 = 0 a2 β 10a + 8aβ 80 = 0 a (a β 10) + 8 (a β 8) = 0 (a β 10) (a + 8) = 0 So, a = 10 or a = β8 Taking a = 10 n = (β 28)/(π + 4) n = (β 28)/(10 + 4) n = (β 28)/14 n = β2 Since n is number of terms, it cannot be negative So, n = β2 is not possible β΄ a = 10 is not possible Taking a = β 8 n = (β 28)/(π + 4) n = (β 28)/(β 8 + 4) n = (β 28)/( β 4) n = 7 So, n = 7