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+1 vote
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in Arithmetic Progression by (30.9k points)
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If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =

A. \(\frac{1}{n}\)

B. \(\frac{n-1}n{}\)

C. \(\frac{n+1}{2n}\)

D . \(\frac{n+1}{n}\)

1 Answer

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Best answer

Correct answer is D. \(\frac{n + 1}{n}\)

Given: the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers. 

To find: The value of k 

Solution: Sum of terms of A.P = n/2 ( 2a + (n-1) d )First n even natural numbers are: 2, 4, 6, 8, ............ 

It forms an AP where first term a = 2 and common difference d = 4 – 2 = 2 

⇒ sum of n even natural terms = \(\frac{n}{2}\)x {2 x 2 + (n–1)2}

= \(\frac{n}{2}\)( 4 + 2n - 2 )

=\(\frac{n}{2}\)( 2 + 2n )

=\(\frac{n}{2}\) × 2 ( 1 + n ) 

= n (n + 1) 

First n odd natural numbers are: 1, 3, 5, 7, ............ 

It forms an AP where first term is a = 1 and the common difference d = 3 – 1 = 2 

Now sum of n terms = \(\frac{n}{2}\){2 x 1 + (n–1)2} 

= \(\frac{n}{2}\)( 2 + 2n - 2 )= \(\frac{n}{2}\)(2n ) 

= n x n

= n2

Now, According to given condition Sum of first n even numbers = k X (Sum of first n odd numbers) 

⇒ n (n + 1) = k x n2 

⇒ k x n = n + 1

⇒ k = \(\frac{(n+1)}{n}\)

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