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Given: The sequence \(a_1, a_2, …, a_n, ….\) is such that \(a_1 = 3, a_2 = 5\), and \( a_n = a_{n-1} – a_{n-2}\) for all integers n ≥3.
Asked: What is the sum of the first \(100\) terms of the sequence?

\(a_1 = 3, a_2 = 5, a_3 = 5-3=2, a_4=2-5=-3\), \(a_5=-3-2=-5,a_6=-5-(-3)=-2;a_7=-2-(-5)=3,a_8=3-(-2)=5\)

Sum of every 6 terms is 0;
100 = 6*16 + 4
Sum of 96 terms = 0
Sum of last 4 terms = 3+5+2-3=7

Sum of first 100 terms = 0+7 = 7

IMO C
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for all integers n≥3; doesn't this mean we have to find sum of all 100 terms that start from A3 i.e A3 to A103
Answer would be diff in that case
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for all integers n≥3; doesn't this mean we have to find sum of all 100 terms that start from A3 i.e A3 to A103
Answer would be diff in that case
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Bhavesh1291997
for all integers n≥3; doesn't this mean we have to find sum of all 100 terms that start from A3 i.e A3 to A103
Answer would be diff in that case
No. “For all integers n ≥ 3” only specifies for which values of n the formula applies. It does not redefine which terms belong to the sequence. The sequence is explicitly defined starting with a1 and a2, and the rule is used to generate a3, a4, and so on. Therefore, “the first 100 terms” means a1 through a100, not a3 through a103.
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