Question

Four different integers from an increasing AP. One of these numbers is equal to sum of the squares of the first three numbers, and then the common difference of the four numbers is
A. 1
B. 2
C. 3
D. -3

Answer 1

Hint: First of all we have to remember one thing is that whenever we have to assume the numbers of arithmetic progression, we always take such that on adding these terms we get minimum numbers of unknowns .For example, if we consider three numbers of A.P like a-d, a, a+d then the sum of these numbers will give only one unknown i.e. a. Using these concepts we will solve the question accordingly.

Complete step-by-step answer:
Given
There are four integers forming an increasing A.P.
Let the four integers be a-d, a, a+d and a+2d
From the question, it is given that the one of the numbers from the AP is equal to the sum of squares of the first three numbers.
Then according to the condition
$ \Rightarrow {\left( {a - d} \right)^2} + {a^2} + {\left( {a + d} \right)^2} = a + 2d$
Expanding the above equation using algebraic identities and simplifying the equation
$
   \Rightarrow {\left( {a - d} \right)^2} + {a^2} + {\left( {a + d} \right)^2} = a + 2d \\
   \Rightarrow {a^2} + {d^2} - 2ad + {a^2} + {a^2} + {d^2} + 2ad = a + 2d \\
   \Rightarrow 3{a^2} + 2{d^2} = a + 2d \\
   \Rightarrow 2{d^2} - 2d + 3{a^2} - a = 0 \\
$
Now, we obtain a quadratic equation in the form of d
As we know the solution quadratic equation $a{x^2} + bx + c = 0$ is given as
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using the above formula
$ \Rightarrow 2{d^2} - 2d + 3{a^2} - a = 0$
$
  d = \dfrac{{2 \pm \sqrt {4 - 4 \times 2 \times \left( {3{a^2} - a} \right)} }}{{2 \times 2}} \\
  d = \dfrac{{1\left[ {1 \pm \sqrt {1 + 2a - 6{a^2}} } \right]}}{2} \\
 $
Since, d is a positive integer
$
  1 + 2a - 6{a^2} > 0 \\
  {a^2} - \dfrac{a}{3} - \dfrac{1}{6} < 0 \\
$
Again, this is a quadratic equation and the solution of this equation can be given using the formula
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Substituting the value of the constants
$
  a = \dfrac{{\dfrac{1}{3} \pm \sqrt {{{\left( {\dfrac{1}{3}} \right)}^2} - 4 \times 1 \times \dfrac{1}{6}} }}{{2 \times 1}} \\
  a = \dfrac{{1 \pm \sqrt 7 }}{6} \\
$
Now form the condition ${a^2} - \dfrac{a}{3} - \dfrac{1}{6} < 0$
The value of a can be represented as
\[
  \left( {a - \dfrac{{1 - \sqrt 7 }}{6}} \right)\left( {a - \dfrac{{1 + \sqrt 7 }}{6}} \right) < 0 \\
  \dfrac{{1 - \sqrt 7 }}{6} < a < \dfrac{{1 + \sqrt 7 }}{6} \\
\]
Since, it is given that a is an integer, therefore a = 0
Now we will find the value of d by substituting the value of a in the equation $d = \dfrac{{1\left[ {1 \pm \sqrt {1 + 2a - 6{a^2}} } \right]}}{2}$
\[
   \Rightarrow d = \dfrac{{1\left[ {1 \pm \sqrt {1 + 2a - 6{a^2}} } \right]}}{2} \\
   \Rightarrow d = \dfrac{{1\left[ {1 \pm \sqrt {1 + 2 \times 0 - 6 \times {0^2}} } \right]}}{2} \\
   \Rightarrow d = 1\,or\,0 \\
 \]
Therefore the four numbers are -1, 0, 1, 2
Hence, the correct option is A.

Note- In order to solve these types of problems, you need to have the basic concept of linear and algebraic equations. When we have to find the first term of an AP or the common difference, the first step is to read the statement and write down the conditions given. When we have to assume a sequence of AP it is better to assume as we mentioned above when we have to eliminate the variables and only one condition is given.