Bunuel wrote:
Sequence P is defined by \(p_n = p_{n-1}+ 3\), \(p_1= 11\), sequence Q is defined as \(q_n=q_{n-1}– 4\), \(q_3= 103\). If \(p_k > q_{k+2}\), what is the smallest value k can take?
A.6
B. 9
C.11
D.14
E.15
p1 = 11
p2 = 11 + 3 = 14
p3 = 14 + 3 = 17
.
.
.
pk = 11 + (k - 1)3 = 8 + 3k
q2 = q3 + 4 = 107
q1 = q2 + 4 = 111
.
.
.
q(k+2) = 111 + (k + 2 - 1)(-4) = 111 - 4k - 4 = 107 - 4k
pk > q(k+2)
—> 8 + 3k > 107 - 4k
—> 7k > 99
—> k > 14.14
k = 15
IMO Option E
Pls Hit kudos if
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