Question

If \((2n+1)+(2n+3)+(2n+5)+….+(2n+47)=5280,\) then what is the value of \(1+2+3+….n?\)

• A. 4851
• B. 1458
• C. 4718
• D. 4378

Answer 1

The Correct Option is A

Given the sum of the arithmetic sequence: \[ (2n + 1) + (2n + 3) + (2n + 5) + \ldots + (2n + 47) = 5280 \] Find the value of \( n \), and then compute: \[ 1 + 2 + 3 + \ldots + n \]

Step 1: Identify the Arithmetic Progression

This is an arithmetic sequence with:

• First term \( a = 2n + 1 \)
• Common difference \( d = 2 \)
• Last term \( l = 2n + 47 \)

 

To find the number of terms \( m \), we use the formula: \[ \text{Last term} = a + (m - 1)d \] Plugging in: \[ 2n + 47 = (2n + 1) + (m - 1) \cdot 2 \Rightarrow 2n + 47 = 2n + 1 + 2m - 2 \Rightarrow 2m = 48 \Rightarrow m = 24 \]

Step 2: Use Sum of AP Formula

Sum of the arithmetic sequence is: \[ S = \frac{m}{2} \cdot (\text{First term} + \text{Last term}) \Rightarrow S = \frac{24}{2} \cdot \left( (2n + 1) + (2n + 47) \right) \Rightarrow 12 \cdot (4n + 48) = 5280 \] Simplifying: \[ 48(n + 12) = 5280 \Rightarrow n + 12 = 110 \Rightarrow n = 98 \]

Step 3: Sum of First \( n \) Natural Numbers

Now that \( n = 98 \), compute: \[ 1 + 2 + 3 + \ldots + 98 = \frac{98 \cdot 99}{2} = 4851 \]

Final Answer: \( \boxed{4851} \)