Exercise Set 8.1
Last updated at June 9, 2026 by Teachoo
Transcript
Ex 8.1, 1 (i) Find the first five terms of the sequence in which the nth term is given by (i) π‘_π=3πβ4 We need to find first five terms i.e. π_π, π_π, π_π, π_π, π_π Given π‘_π=3πβ4 Putting n = 1 in π_π π‘_1=3(1)β4 = 3 β 4 = β1 Putting n = 2 in π_π π‘_2=3(2)β4 = 6 β 4 = 2 Putting n = 3 in π_π π‘_3=3(3)β4 = 9 β 4 = 5 Putting n = 4 in π_π π‘_4=3(4)β4 = 12 β 4 = 8 Putting n = 5 in π_π π‘_5=3(5)β4 = 15 β 4 = 11 Thus, first 5 terms are β1, 2, 5, 8, 11 Ex 8.1, 1 (ii) Find the first five terms of the sequence in which the nth term is given by (ii) π‘_π=2β5π We need to find first five terms i.e. π_π, π_π, π_π, π_π, π_π Given π‘_π=2β5π Putting n = 1 in π_π π‘_1=2β5(1) = 2 β 5 = β3 Putting n = 2 in π_π π‘_2=2β5(2) = 2 β 10 = β8 Putting n = 3 in π_π π‘_3=2β5(3) = 2 β 15 = β13 Putting n = 4 in π_π π‘_4=2β5(4) = 2 β 20 = β18 Putting n = 5 in π_π π‘_5=2β5(5) = 2 β 25 = β23 Thus, first 5 terms are β3, β8, β13, β18, β23 Ex 8.1, 1 (iii) Find the first five terms of the sequence in which the nth term is given by (iii) π‘_π=π^2β2π+3 for πβ₯1. We need to find first five terms i.e. π_π, π_π, π_π, π_π, π_π Given π‘_π=π^2β2π+3 Putting n = 1 in π_π π_π=1^2β2(1)+3 = 1 β 2 + 3 = 4 β 2 = 2 Putting n = 2 in π_π π_π=2^2β2(2)+3 = 4 β 4 + 3 = 0 + 3 = 3 Putting n = 3 in π_π π_π=3^2β2(3)+3 = 9 β 6 + 3 = 12 β 6 = 6 Putting n = 4 in π_π π_π=4^2β2(4)+3 = 16 β 8 + 3 = 19 β 8 = 11 Putting n = 5 in π_π π_π=5^2β2(5)+3 = 25 β 10 + 3 = 28 β 10 = 18 Thus, first 5 terms are 2, 3, 6, 11, 18