Consider the sequence 2017, 2018, 2019, . . . . ., an such that an=an-3 +an-2- an-1 for all n 4 . i.e., 4th term is 2017 +2018 -2019 and so on, then a1990 is
Answers
Given : sequence 2017, 2018, 2019, . . . . .,
aₙ = aₙ₋₃ + aₙ₋₂ - aₙ₋₁
To Find : a₁₉₉₀
Solution:
a₁ = 2017
a₂ = 2018
a₃ = 2019
a₄ = a₁ + a₂ - a₃ = 2017 + 2018 - 2019 = 2016
a₅ = a₂ + a₃ - a₄ = 2018 + 2019 - 2016 = 2021
a₆ = a₃ + a₄ - a₅ = 2019 + 2016 - 2021 = 2014
a₇ = a₄ + a₅ - a₆ = 2016 + 2021 - 2014 = 2023
a₈ = a₅ + a₆ - a₇ = 2021 + 2014 - 2023 = 2012
Notice carefully
2017, 2019 , 2021 , 2023....................
2018 , 2016 , 2014 , 2012 ,.................
a₁₉₉₀ = 995th term in 2018 , 2016 , 2014 , 2012 ,.................
a = 2018
d = -2
n = 995
2018 + (995 - 1) (-2)
= 2018 -1988
= 30
a₁₉₉₀ = 30
or another logic we can perceive
aₙ - n = 2016 for n = odd ( 2017 - 1 = 2019 - 3= 2021 - 5 = 2023 - 7 ......)
aₙ +n = 2020 for n = even ( 2018 + 2 = 2016 + 4 = 2014 + 6 = 2012 + 8 ....)
a₁₉₉₀ => n = 1990 even
Hence a₁₉₉₀ + 1990 = 2020
=> a₁₉₉₀ = 30
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