If a, b, c, d, are distinct integers in AP such that d=a^2+b^2+c^2, then the sum a+b+c+d equals

step 1 Hi, we're giving here ABCD are distinct integers in AP. So that is the relation is given. Then a

If a, b, c, d, are distinct integers in AP such that...

Key Concepts

Arithmetic Progression

An arithmetic progression is a sequence of numbers where each term after the first is obtained by adding a constant difference to the preceding term. This concept allows for a simple algebraic representation of sequence terms, which is essential when establishing relationships between them.

Algebraic Representation of Sequence Terms

Representing the terms of an arithmetic progression using expressions involving the first term and the common difference simplifies the formulation of equations. This technique is fundamental in translating word problems into solvable algebraic equations.

Formulating and Solving Quadratic Equations

In problems where conditions relate various terms of a sequence, the resulting equation may be quadratic. The process of forming and solving quadratic equations is crucial for determining the values of variables that satisfy these conditions.

Constraints in Integer Sequences

When dealing with integer sequences, additional constraints such as distinctness impose further restrictions. Ensuring that the numbers satisfy all given conditions, including being distinct integers, is key to arriving at the correct solution.

Transcript

00:02 Hi, we're giving here abcd are distinct integers in ap.

00:07 So that is the relation is given.

00:09 Then a plus b plus c plus d equals.

00:13 So how we can approach this problem? we find out d in terms of common difference in the first term, a, b, c also.

00:20 We get a relation in k and then we take common difference as k, let's say.

00:26 We had a relation in k in that we solve the inequality for discriminant.

00:32 And then we get some values and solve for a plus b.

00:34 Plus c plus d let's go for this so cost just let's say we have the common difference is k common difference bk so if we just work on this we have d equals a square plus b square plus c square so d would be a plus three k right that's a fourth term equals a square plus a plus k whole square plus a plus two k whole square we'll find this we have have a plus 3k so a square plus a square plus two a k plus a square plus a square plus four k square plus four a k this will work on this so we have a k square four k square that's giving five k square next we get three k times uh next we have uh okay so four a k plus two a k that's giving six a k next we have a next we have a 3k here, negative 3k, next plus a square, a square, a square, three a square, next negative a, equal zero.

02:04 We get 5, k square, positive, 3k, 2a, 2a ,000, positive, 3a square negative a, equal 0.

02:18 Now, k is real, it's an ap, so the k is real.

02:22 It's a quality equation in k here, we have k -square.

02:24 K is real, right? it's an ap.

02:27 K is real.

02:28 Now, k is real we have.

02:30 So, we can say discriminant, is squared and equal to zero.

02:34 For real and distinct, k is real and distinct.

02:36 Discriminate is equal to zero.

02:38 Let's find discriminant.

02:39 So that is given as b square, nine, two and negative one, whole square, negative, four times, five times three a square negative eight, a squared than equal to let's work on this.

02:56 You get 9 times 4a square, positive 1 negative 4a.

03:01 Negative 20, 3a squared negative a squared squared than equal to 0.

03:08 Let's work on this.

03:09 We get 36a squared, negative 36a, positive 9, negative 60 a square, negative 208, squared.

03:20 Simplifying further.

03:23 So we have 36a square, then we have negative 60 that gives us.

03:29 Negative 24 a square then we have negative 36a just coming out to a positive 28 here if it's positive that will be negative 16a and then it's come out to positive 9 greater than equal to zero multi -line both sides with negative 1 we get 24 a square positive 16a negative 9 is less than equal to 0 now in the version of a square to be 1 we get a square plus 16 over 24 that is 2 over 3a negative 3 over 8 is less than equal to 0 now we apply a permitting square here a square plus 2 over 3a plus 1 over 3 square negative 1 over 9 negative 3 over 8 is less than equal to 0 we have a plus 1 over 3 whole square is less than equal to 3 over 8 plus 1 over 9 you are simplifying this so equal we have .7 plus 8 over 72 that equals 35 over 72 or that is 70 over 144.

04:55 So, we have got a plus 1 over 3 whole square is less than equal to 70 over 144.

05:03 So, go here.

05:05 And the next step is next year we have this language in this, so we can say it's going to have to be negative root 70 12 is less than equal to a plus 1 over 3 doesn't equal to root 70 over must the range we've got here we're doing negative 1 over 3 here negative 1 over 3 here and negative 1 over 3 here but solve for 8 so we get tass negative root 70 over 12 negative 1 over 3 that gives negative 4 negative root 70 over 12 less than equal to a less than equal to we have root 70 and negative we have 4 over 12.

05:56 If we work on this now, so we have root 70.

06:00 Route 70 is lying between we have 8 and 9.

06:06 It's going to be 8 point something, right? so therefore it's going to be negative 4 and then negative 8 point something.

06:14 So that will be something something like to be we have this part.

06:20 It's coming have to be less than negative one.

06:26 Because this is lying between 8 and 9, this is less than negative 1.

06:30 So therefore, and this will be here, route 78 point something, this will be here just greater than 0 and less than 1, right? greater than 0 and less than 1, this could be lying between 0 and 1.

06:47 This line between negative 1 and 0 and 1.

06:49 This line between negative 1 and 0.

06:50 So less than negative one and greater than zero.

06:55 So therefore, we can say here, or to make it simplify what we can say in the way, we can just apply to simplify this not required.

07:08 It's less than negative one.

07:10 So that means so we have the range here basically a.

07:14 A will include.

07:15 So negative one, we can say like this, negative one and zero are included for a.

07:21 We can say like this are included for a.

07:24 Now, the ap consists of integers we have, so a could be negative 1 or 0...