Four different integers form an increasing A.P. If one of...

Key Concepts

Arithmetic Progression

An arithmetic progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed constant, known as the common difference, to the preceding term. This concept helps in forming relationships between terms and setting up equations that describe the behavior of the sequence.

Representation of Terms in an AP

In problems involving arithmetic progressions, a common technique is to denote the terms algebraically (for example, as a, a + d, a + 2d, a + 3d), which allows one to express conditions or constraints given in the problem in terms of these variables. This representation is crucial for forming equations that encapsulate the problem's requirements.

Solving Equations with Integer Constraints

When a problem specifies that the elements of a sequence are integers, it becomes a challenge in number theory and algebra to find solutions that satisfy both the algebraic equation and the discrete nature of integers. Such problems often lead to Diophantine equations, where the strategy includes testing for potential integer values that meet all given conditions.

Sum of Squares

The sum of squares refers to adding together the squares of individual numbers. In various algebra problems, especially those blending sequence properties with quadratic expressions, this concept is used to form additional equations that combine linear and quadratic components, providing a deeper layer of complexity in linking terms of a sequence.

Transcript

00:01 In this question, four different integers form an increasing ap if one of these numbers is equal to the sum of the squares of the other three numbers, then the number is.

00:19 So in this question, let the numbers be, let the numbers be.

00:31 Let the numbers be.

00:42 A minus d a a plus d a plus d a plus 2d a plus 2d so these are the four numbers then according to our hypothesis a minus according to hypothesis squares of small three numbers sum of squares of small three numbers is equal to the fourth number that is sum of a minus d plus a sorry sum of a plus a minus d whole square plus a square plus a my a plus d whole square is equal to a plus two d so that is a minus d whole square plus a square plus a square plus a plus d whole square is equal to a plus two d okay okay, now we can rearrange the equation.

02:03 So we will get 2d square to 2d square minus 2d plus 3a square minus 2d plus 3a square minus a is equal to 0.

02:23 That is 2d2d plus 3a square minus a is equal to 0 now since d is a positive integer since d is a positive integer since d is a positive integer you know since d is a positive integer now we can before that we want to convert this into the you want to find a solution of this equation we want to find a solution of this equation so d is equal to d is equal to 1 by 2 1 plus or minus root of 1 square plus 2a minus 6 a square this is the solution of the equation d now since d is a positive integer since this positive integer this term this term must be greater than 0 that is 1 square plus 2a minus 6a square must be greater than 0 so 1 plus 2a minus 6a square which is greater than 0 now rearrange this equation by putting minus 6a square to other side and these two into other side so we and also by dividing the whole term by six we will get a square minus a by three minus one by six less than zero okay by solving this we will get that now a minus a sorry as a minus a minus 1 minus root 7 by 6 into a minus a minus 1 the application into in this form okay now here we know we will get that 1 minus root 7 by 6 1 minus root 7 by 6 is less than a or is greater than a or less than 1 plus root 7 by 6 so a will a must be less than 1 plus root 7 by 6 and greater greater than 1 minus root 7 by 6 now since a is an integer here we know since a sorry since a is an integer since a is an integer is an integer we will get that this is this term is less than zero since is a negative term and this is a positive term so here we will get that a is equal to 0 since a is integer since a is a is equal to 0 okay now now then d then d is equal to zero in the equation then we will get d is equal to half into half into one plus plus or minus one which is equal to 1 or 0.

07:17 We substituted value of a in the above equation, then we will get d is equal to half into 1 plus or minus 1, which is equal to 1 or 0.

07:29 Here also, in the above equation, we know that d is a positive integer.

07:40 We here see that d is a positive integer...

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