The fourth and fifth term of a sequence {tn}n≥1 are 4 and 5 respectively and the nth term is given as tn = 2tn-1 - tn-2, n 3 (n∈ N). Then the sum to first 2009 terms isa)2019045b)2013021 c)2017036d)20180 40Correct answer is option 'A'. Can you explain this answer?

Question

The fourth and fifth term of a sequence {tn}n&ge;1 are 4 and 5 respectively and the nth term is given as tn = 2tn-1 - tn-2, n > 3 (n&isin; N). Then the sum to first 2009 terms isa)2019045b)2013021 c)2017036d)20180 40Correct answer is option 'A'. Can you explain this answer?

Accepted Answer

Thus {a_n} is a constant sequence
a₅ = t₅ − t₄ = 1
Now a₄ = t₄ − t₃ ⇒ 1 = 4 − t₃ ⇒ t₃ = 3
Similarly t₂ = 2 , t₁ = 1
Thus {t_n} is an A.P with r = 1 and common difference 1

Answer

The fourth and fifth terms of the sequence can be represented as t4 and t5.

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