a_(1 ) , a_2 , a_(3 ) , .... , a_n is a sequence satifies that a_(n+2) =a_(n+1) −a_n for n ≥ 1. suppose the sum of the first 999 terms = 1003 and the sum of the first 1003 terms = −999 find the sum of the first 2002 terms.


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Question Number 192597 by York12 last updated on 22/May/23

$a_{1}, a_{2}, a_{3}, . . . ., a_{n} i s a s e q u e n c e s a t i f i e s t h a t$ $a_{n + 2} = a_{n + 1} - a_{n} f o r n \geq 1 . s u p p o s e t h e s u m$ $o f t h e f i r s t 999 t e r m s = 1003 a n d t h e s u m$ $o f t h e f i r s t 1003 t e r m s = - 999 f i n d t h e$ $s u m o f t h e f i r s t 2002 t e r m s .$
	Answered by MM42 last updated on 22/May/23

	$a_{4} = a_{3} - a_{2} = - a_{1} & a_{5} = a_{4} - a_{3} = - a_{2} & a_{6} = a_{5} - a_{4} = - a_{3}$ $\Rightarrow a_{6 k + 1} = a_{1} & a_{6 k + 2} = a_{2} & a_{6 k + 3} = a_{3}$ $a_{6 k + 4} = - a_{1} & a_{6 k + 5} = - a_{2} & a_{6 k} = - a_{3}$ $999 = 6 \times 166 + 3 \Rightarrow S_{999} = a_{1} + a_{2} + a_{3} = 2 a_{2} = 1003 \Rightarrow a_{2} = 501.5$ $1003 = 6 \times 167 + 1 \Rightarrow S_{1003} = a_{1} \Rightarrow a_{1} = - 999$ $\Rightarrow a_{3} = 1500.5$ $2002 = 6 \times 336 + 4 \Rightarrow S_{2002} = a_{1} + a_{2} + a_{3} + a_{4} = a_{2} + a_{3} = 2002$
	Commented by York12 last updated on 22/May/23

	$a_{2} = 501.5$ $s_{2002} = 999 + 2 \times 501.5 = 2002$
	Commented by York12 last updated on 22/May/23

	$a_{4} = a_{3} - a_{2} = - a_{1} & a_{5} = a_{4} - a_{3} = - a_{2} & a_{6} = a_{5} - a_{4} = - a_{3}$ $\Rightarrow a_{6 k + 1} = a_{1} & a_{6 k + 2} = a_{2} & a_{6 k + 3} = a_{3}$ $a_{6 k + 4} = - a_{1} & a_{6 k + 5} = - a_{2} & a_{6 k} = - a_{3}$ $999 = 6 \times 166 + 3 \Rightarrow S_{999} = a_{1} + a_{2} + a_{3} = 2 a_{2} = 1003 \Rightarrow a_{2} = 501.5$ $1003 = 6 \times 167 + 1 \Rightarrow S_{1003} = a_{1} \Rightarrow a_{1} = - 999$ $⇛ a_{3} = 1500.5$ $2002 = 6 \times 336 + 4 \Rightarrow S_{2002} = a_{1} + a_{2} + a_{3} + a_{4} = a_{2} + a_{3} = 2002$
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