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Question: If \[a,b,c,d\] are distinct integers form an increasing A.P. such that \[d = {a^2} + {b^2} + {c^2}\]...

If a,b,c,da,b,c,d are distinct integers form an increasing A.P. such that d=a2+b2+c2d = {a^2} + {b^2} + {c^2}, then find the value of a+b+c+da + b + c + d

Explanation

Solution

Here, we have to find the sum of all distinct integers which are in increasing A.P. First, we will assign the integers as the general series of A.P., then we have to find the value of integers by solving the quadratic equation formed from the condition. So, we will get the value of distinct integers after which we have to add all the values of integers in A.P. Arithmetic Progression (AP) is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.

Formula used:
We will use the following formulas:
1.The A.P. can be written in terms of common difference as follows: a−d,a,a+d,a+2da - d,a,{\rm{ }}a{\rm{ }} + {\rm{ }}d,{\rm{ }}a{\rm{ }} + {\rm{ }}2d.
2.The square of sum of numbers is given by the formula (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab.
3.The square of difference of numbers is given by the formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab.
4.Roots of a quadratic equation is given by the formula x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}.

Complete step-by-step answer:
We are given that a,b,c,da,b,c,d are distinct integers form an increasing A.P.
The A.P. can be written in terms of common difference as a−d,a,a+d,a+2da - d,a,{\rm{ }}a{\rm{ }} + {\rm{ }}d,{\rm{ }}a{\rm{ }} + {\rm{ }}2d.
So, let us consider a=m−n,b=m,c=m+n,d=m+2na = m - n,b = m,c = m + n,d = m + 2n
Here nn is a common difference
Since, d=a2+b2+c2d = {a^2} + {b^2} + {c^2}, we have
Substituting the values of a,b,c,da,b,c,d, we have
⇒m+2n=(m−n)2+m2+(m+n)2\Rightarrow m + 2n = {\left( {m - n} \right)^2} + {m^2} + {\left( {m + n} \right)^2}
The square of sum of numbers is given by the formula (a+b)2=a2+b2+2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab
The square of difference of numbers is given by the formula (a−b)2=a2+b2−2ab{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab ⇒m+2n=m2+n2−2mn+m2+m2+n2+2mn \Rightarrow m + 2n = {m^2} + {n^2} - 2mn + {m^2} + {m^2} + {n^2} + 2mn
Adding and Subtracting the like terms, we have
⇒m+2n=3m2+2n2\Rightarrow m + 2n = 3{m^2} + 2{n^2}
Rewriting the equation, we have
⇒3m2+2n2−2n−m=0\Rightarrow 3{m^2} + 2{n^2} - 2n - m = 0
⇒2n2−2n+(3m2−m)=0\Rightarrow 2{n^2} - 2n + (3{m^2} - m) = 0
Now, we have to solve for mm, we have
If the roots are real, b2−4ac≥0{b^2} - 4ac \ge 0
Since nn is real, we have
⇒4−8(3m2−m)≥0\Rightarrow 4 - 8(3{m^2} - m) \ge 0
Multiplying the terms, we have
⇒4−24m2+8m≥0\Rightarrow 4 - 24{m^2} + 8m \ge 0
Rewriting the equation, we have
⇒24m2−8m−4≤0\Rightarrow 24{m^2} - 8m - 4 \le 0
Dividing by 44on both the sides, we have
⇒6m2−2m−1≤0\Rightarrow 6{m^2} - 2m - 1 \le 0
Roots of a quadratic equation is given by the formula x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}. Therefore,
Roots of a quadratic equation=2±2812 = \dfrac{{2 \pm \sqrt {28} }}{{12}}
⇒\Rightarrow Roots of a quadratic equation=2±2×2×712 = \dfrac{{2 \pm \sqrt {2 \times 2 \times 7} }}{{12}}
⇒\Rightarrow Roots of a quadratic equation=2±2712 = \dfrac{{2 \pm 2\sqrt 7 }}{{12}}
Simplifying the expression, we get
⇒\Rightarrow Roots of a quadratic equation=2(1±7)12 = \dfrac{{2\left( {1 \pm \sqrt 7 } \right)}}{{12}}
⇒\Rightarrow Roots of a quadratic equation=(1±7)6 = \dfrac{{\left( {1 \pm \sqrt 7 } \right)}}{6}
⇒\Rightarrow m∈[1−76,1+76]m \in \left[ {\dfrac{{1 - \sqrt 7 }}{6},\dfrac{{1 + \sqrt 7 }}{6}} \right]
As mm is Integer, m=0m = 0 (only positive integer)
⇒\Rightarrow 2n2−2n+0=02{n^2} - 2n + 0 = 0
⇒\Rightarrow n=0(or)1n = 0\left( {or} \right)1
But common differences cannot be 00 as a,b,c,da,b,c,d are distinct.
⇒m=0;n=1\Rightarrow m = 0;n = 1
Substituting the values of m,nm,n to find the values of a,b,c,da,b,c,d
⇒a=m−n=0−1=−1\Rightarrow a = m - n = 0 - 1 = - 1;
⇒b=m=0\Rightarrow b = m = 0;
⇒c=m+n=0+1=1\Rightarrow c = m + n = 0 + 1 = 1;
⇒d=m+2n=0+2(1)=2\Rightarrow d = m + 2n = 0 + 2(1) = 2
Now, we have to find the sum of all distinct integers.
⇒a+b+c+d=−1+0+1+2\Rightarrow a + b + c + d = - 1 + 0 + 1 + 2
⇒a+b+c+d=2\Rightarrow a + b + c + d = 2

Note: We have to know the properties of A.P. such as if the same number is added or subtracted from each term of an A.P, then the resulting terms in the sequence are also in A.P with the same common difference. We can check if the values we found are in A.P. by the property three numbers x,yx,y and zz are in an A.P if 2y=x+z2y = x + z. If we select terms in the regular interval from an A.P, these selected terms will also be in AP