Question

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?
9
10
12
14

Answer 1

12

It is the correct answer

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Answer 2

Answer:

Step-by-step explanation:

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

S2n = 2n2

2

n

2

[2 * 1+(2n −1)4]

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

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

Sn = n2

n

2

[2 * 56+(n −1)2]

Sn = n2

n

2

[112+2n −2] = 2n2

2

n

2

(110+2n)

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 > n2

n

2

(110+2n)

16n2 – 4n > 110n + 2n2

14n2 > 114n

7n > 57

n > 577

57

7

The smallest value n can take = 9.