Question

If \[ 1+3+5+\cdots+l_1=1521 \] and \[ 2+4+6+\cdots+l_2=1722, \] then $l_1+l_2=$

• A. 160
• B. 159
• C. 80
• D. 79

Hint:

Memorize: \[ 1+3+5+\cdots +(2n-1)=n^2 \] and \[ 2+4+6+\cdots+2n=n(n+1). \] These formulas frequently appear in objective examinations.

Answer 1

The Correct Option is B

Concept: The sum of the first $n$ odd natural numbers is \[ 1+3+5+\cdots +(2n-1)=n^2. \] The sum of the first $n$ even natural numbers is \[ 2+4+6+\cdots+2n=n(n+1). \]

Step 1: Determine $l_1$. Given \[ 1+3+5+\cdots+l_1=1521. \] Since the sum of first $n$ odd numbers is $n^2$, \[ n^2=1521. \] \[ n=39. \] Hence \[ l_1=2n-1=77. \]

Step 2: Determine $l_2$. Given \[ 2+4+6+\cdots+l_2=1722. \] Using \[ n(n+1)=1722. \] \[ 41\times42=1722. \] Thus \[ n=41. \] Hence \[ l_2=2n=82. \]

Step 3: Find the required sum. \[ l_1+l_2 = 77+82 = 159. \] Therefore, \[ \boxed{159}. \]