# Arithmetic and Geometric Progressions

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## Arithmetic Progression

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?
1. 6
2. 11
3. 14
4. 15

Choice D. 15

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## Detailed Solution

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

8 + 3k > 107 – 4k

7k > 99

k > 99/7

k has to be an integer, so smallest value k can take is 15.

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## More questions from Progressions

With some simple but very powerful ideas, one can cut down on a lot of working when it comes to progressions. For example, anchoring a progression around its middle term can be very useful. Reinforce these ideas with these questions.