First slide

Arithmetic progression

Question

Four different integers form an increasing A.P. If one of these numbers is equal to the sum of the squares of the other three numbers, then the numbers are

Moderate

A

$- 2, - 1, 0, 1$
B

$0, 1, 2, 3$
C

$- 1, 0, 1, 2$
D

None of these

Solution

Let the numbers be $a - d, a, a + d, a + 2 d$
where, $a, d \in Z and d > 0$
Given: $(a - d)^{2} + a^{2} + (a + d)^{2} = a + 2 d$
$\Rightarrow 2 d^{2} - 2 d + 3 a^{2} - a = 0$
$∴ d = \frac{1}{2} [1 \pm \sqrt{(1 + 2 a - 6 a^{2})}]$
Since d is positive integer,
$\begin{array}{ll} ∴ & 1 + 2 a - 6 a^{2} > 0 \\ \Rightarrow & (\frac{1 - \sqrt{7}}{6}) < a < (\frac{1 + \sqrt{7}}{6}) \end{array}$
Since a is an integer,
$∴ a = 0$ ,
then $d = \frac{1}{2} [1 \pm 1] = 1 or 0$ Since $d > 0$
$∴ d = 1 .$
Hence, the numbers are $- 1, 0, 1, 2$ .

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