Math, asked by Kalden7996, 1 year ago

Find the consecutive odd natural numbers some of whose square is 970

Answers

Answered by Anonymous
1

Answer:

21, 23

( 21² + 23² = 441 + 529 = 970 )

Step-by-step explanation:

Let the two consecutive odd natural numbers be 2n-1 and 2n+1, where n≥1 is an integers.

The sum of their squares is 970

=> (2n-1)² + (2n+1)² = 970

=> 4n² - 4n + 1  +  4n² + 4n + 1 = 970

=> 8n² = 968

=> n² = 121

=> n = 11.

Therefore the two consecutive odd natural numbers are

2n-1 = 21  and  2n+1 = 23.

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