# Arithmetic and Geometric Progressions

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## Arithmetic Progression

The sum of 2n terms of A.P. {1, 5, 9, 13…..} is greater than sum of n terms of A.P. = {56, 58, 60..…}. What is the smallest value n can take?
1. 9
2. 10
3. 12
4. 14

Choice A. 40000

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## Detailed Solution

First A.P., a = 1, d = 4

S2n = 2n/2 [2 x 1+(2n −1)4]

S2n = n(2 + 8n – 4) = n(8n – 2) = 8n2 – 2n

For the second AP, a = 56, d = 2

Sn = n/2 [2 x 56+(n −1)2]

Sn = n/2 [112+2n −2] = n/2 (110+2n)

The sum of 2n terms of AP {1, 5, 9, 13, …..} is greater than sum of n terms of A.P. = {56, 58, 60, …}

8n2 – 2n > n/2 (110+2n)

16n2 – 4n > 110n + 2n2

14n2 > 114n

7n > 57

n > 57/7

The smallest value n can take = 9. Answer choice (a).

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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.