If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers,

Question:

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) $\frac{n+1}{2 n}$

(d) $\frac{n+1}{n}$

Solution:

(d) $\frac{n+1}{n}$

Given:

Sum of the even natural numbers = ×">× Sum of the odd natural numbers

$\frac{n}{2}\{2 a+(n-1) d\}=k \times \frac{n}{2}\{2 a+(n-1) d\}$

$\Rightarrow\{2 \times 2+(n-1) 2\}=k \times\{2 \times 1+(n-1) 2\}$

$\Rightarrow \frac{4+(n-1) 2}{2+(n-1) 2}=k$

$\Rightarrow \frac{n+1}{n}=k$

 

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now