Question

a. express 196 as the sum of 14'odd number


b. write a Pythagorean triplet whose one member is 18

Answer 1

a.1+3+5+7+9+11+13+15+17+19+21+23+25+27
b.2x,x×x-1,x×x+1
2x=18
x=9
9×9-1=80
9×9+1=82
Therefore,Pythagorean triplet of 18 is 9,80,82

Answer 2

a.

The sum of first n consecutive odd numbers is n².

Here 196 = 14².

∴ 196 is the sum of first 14 odd numbers.

nth odd number = 2n - 1

∴ 14th odd number = 2 x 14 - 1 = 28 - 1 = 27.

∴ 1 + 3 + 5 + ...... + 25 + 27 = 196

b.

If a² + b² = c², then a = m² - n², b = 2mn and c = m² + n².

We can consider 18 as b = 2mn.

2mn = 18

mn = 18 / 2 = 9

9 = 9 x 1

∴ m = 9 and n = 1 can be considered.

a = 9² - 1² = 81 - 1 = 80

b = 2 x 9 x 1 = 18

c = 9² + 1² = 81 + 1 = 82

∴ 80, 18 and 82 is a Pythagorean triplet.

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