# Ex.6.1 Q8 Squares and Square Roots Solutions - NCERT Maths Class 8

Go back to 'Ex.6.1'

## Question

(i) Express \(49\) as the sum of \(7\) odd numbers.

(ii) Express \(121\) as the sum of \(11\) odd numbers.

## Text Solution

**What is known?**

Sum of \(7\) odd number is \(49\)

Sum of \(11\) odd number is \(121\)

**What is uknown?**

Express \(49\) as the sum of \(7 \) odd numbers and \(121\) as the sum of 11 odd numbers.

**Reasoning:**

The sum of successive odd natural numbers is \(n^2\).

**Steps:**

(i) \(49 = {\left( 7 \right)^2}\)

Therefore ,\(49\) is the sum of first \(7\) odd natural numbers

\(49 = 1 + 3 + 5 + 7 + 9 + 11 + 13\)

(ii) \(121={\left(11 \right)^2}\)

Therefore ,\(121\) is the sum of first \(11\) odd natural numbers

\(121 = \left[ \begin{array}{l} 1 + 3 + 5 + 7 + 9 + 11 + \\13 + 15 + 17 + 19 + 21 \\ \end{array} \right]\)