Answer
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Hint- The series of numbers that are more than 1 from its preceding number id known as consecutive numbers. The consecutive number series can be even as well as odd depending on the first term of the series and the difference between the numbers. If the difference between the two consecutive numbers is 2 and the starting term is an odd integer then, it is known as the series of the odd consecutive numbers, whereas if the difference between the two consecutive numbers is 2 and the starting term is an even integer, then it is known to be an even consecutive number series.
Complete step by step solution:
Consider the first number to be $x$ so that the consecutive number will be $ x + 1 $.
Now, according to the question, the sum of the squares $ x $ and $ x + 1 $ is 365. So, the equation is written as: $ {x^2} + {(x + 1)^2} = 365 $.
Solve the equation $ {x^2} + {(x + 1)^2} = 365 $ by using the splitting the middle term method to determine the value of $x$ as:
$
{x^2} + {(x + 1)^2} = 365 \\
{x^2} + {x^2} + 2x + 1 = 365 \\
2{x^2} + 2x + 1 - 365 = 0 \\
2{x^2} + 2x - 364 = 0 \\
{x^2} + x - 182 = 0 \\
$
Now, split the middle term into two parts such that the sum will be the coefficient of $ x $ and the product will be the product of the coefficient of $ {x^2} $ and $ {x^0} $.
$
{x^2} + 14x - 13x - 182 = 0 \\
x(x + 14) - 13(x + 14) = 0 \\
(x + 14)(x - 13) = 0 \\
x = 13, - 14 \\
$
As the question is asking for the positive consecutive numbers only, so drop $ x = - 14 $ and the desired value of $ x $ is 13.
Now, substitute $ x = 13 $ in the function $ x + 1 $ so as to determine the value of the second number as: $ x + 1 = 13 + 1 = 14 $.
Hence, 13 and 14 are two consecutive positive integers, the sum of whose square is 365.
Note: It is interesting to note here that for a series of even or odd consecutive numbers, all the terms present in the series are even and odd, respectively.
1,2,3,4,… is a series of general, consecutive numbers.
2,4,6,8,… is a series of even consecutive numbers.
1,3,5,7,… is a series of odd consecutive numbers.
Complete step by step solution:
Consider the first number to be $x$ so that the consecutive number will be $ x + 1 $.
Now, according to the question, the sum of the squares $ x $ and $ x + 1 $ is 365. So, the equation is written as: $ {x^2} + {(x + 1)^2} = 365 $.
Solve the equation $ {x^2} + {(x + 1)^2} = 365 $ by using the splitting the middle term method to determine the value of $x$ as:
$
{x^2} + {(x + 1)^2} = 365 \\
{x^2} + {x^2} + 2x + 1 = 365 \\
2{x^2} + 2x + 1 - 365 = 0 \\
2{x^2} + 2x - 364 = 0 \\
{x^2} + x - 182 = 0 \\
$
Now, split the middle term into two parts such that the sum will be the coefficient of $ x $ and the product will be the product of the coefficient of $ {x^2} $ and $ {x^0} $.
$
{x^2} + 14x - 13x - 182 = 0 \\
x(x + 14) - 13(x + 14) = 0 \\
(x + 14)(x - 13) = 0 \\
x = 13, - 14 \\
$
As the question is asking for the positive consecutive numbers only, so drop $ x = - 14 $ and the desired value of $ x $ is 13.
Now, substitute $ x = 13 $ in the function $ x + 1 $ so as to determine the value of the second number as: $ x + 1 = 13 + 1 = 14 $.
Hence, 13 and 14 are two consecutive positive integers, the sum of whose square is 365.
Note: It is interesting to note here that for a series of even or odd consecutive numbers, all the terms present in the series are even and odd, respectively.
1,2,3,4,… is a series of general, consecutive numbers.
2,4,6,8,… is a series of even consecutive numbers.
1,3,5,7,… is a series of odd consecutive numbers.
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