What is the square root of the sum of first 36 odd natural numbers?

Options

A. 6
B. 36
C. 64
D. 72

Answer

To find the square root of the sum of the first 36 odd natural numbers, we can follow these steps:

### Step 1: Identify the first 36 odd natural numbers
The first 36 odd natural numbers are:
1, 3, 5, 7, ..., up to the 36th term.

### Step 2: Recognize the pattern
These numbers form an arithmetic progression (AP) where:
- The first term \( A = 1 \)
- The common difference \( D = 2 \)

### Step 3: Find the last term
To find the last term (LT) of the first 36 odd natural numbers, we can use the formula for the nth term of an AP:
\[ LT = A + (N - 1) \times D \]
Where:
- \( N = 36 \) (the number of terms)

Substituting the values:
\[ LT = 1 + (36 - 1) \times 2 = 1 + 35 \times 2 = 1 + 70 = 71 \]

### Step 4: Calculate the sum of the first 36 odd natural numbers
The sum \( S \) of the first \( N \) terms of an AP can be calculated using the formula:
\[ S = \frac{N}{2} \times (A + LT) \]
Substituting the values:
\[ S = \frac{36}{2} \times (1 + 71) = 18 \times 72 = 1296 \]

### Step 5: Find the square root of the sum
Now, we need to find the square root of the sum:
\[ \sqrt{S} = \sqrt{1296} \]
Calculating the square root:
\[ \sqrt{1296} = 36 \]

### Final Answer
The square root of the sum of the first 36 odd natural numbers is **36**.

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