Sequence P is an arithmetic progression with initial term 11 and common difference 3. Sequence Q is an arithmetic progression with third term 103 and common difference -4. To find the smallest value of k where the term pk is greater than qk+2, the expressions for the kth terms of P and Q are set up and solved for k. The smallest integer value of k that satisfies this is 15.

1. Progressions Q5

2. Qn: Arithmetic Progression
(a) 6 (b) 11
(c) 14 (d) 15
Sequence P is defined by pn = pn–1 + 3, p1 = 11, Sequence Q is defined
as qn = qn–1 – 4, q3 = 103. If pk > qk+2, what is the smallest value k can
take?

3. Soln: Arithmetic Progression
Sequence P is an A.P. with a = 11, and common difference 3.
So, Pk = 11 + (k – 1)3.
Sequence Q is an A.P. with third term 103 and common difference – 4.
t3 = a + 2d
103 = a + 2 (– 4) or a = 111
qk + 2 = 111 + (k +1) (– 4)
qk + 2 = 111 – 4k – 4 = 107 – 4k
pk > qk+2
11 + (k–1)3 > 107 – 4k
Sequence P is defined by pn = pn–1 + 3, p1 = 11, Sequence Q is defined
as qn = qn–1 – 4, q3 = 103. If pk > qk+2, what is the smallest value k can
take?

4. Soln: Arithmetic Progression
8 + 3k > 107 – 4k
7k > 99
k >
99
7
k has to be an integer, so smallest value k can take is 15.
Answer choice (d)
Sequence P is defined by pn = pn–1 + 3, p1 = 11, Sequence Q is defined
as qn = qn–1 – 4, q3 = 103. If pk > qk+2, what is the smallest value k can
take?