Question: The sum of 3 numbers in an AS is 12 and the sum of their squares is 66. Find the numbers.

Solution:

Let the three numbers in AS be (a-d), a and (a +d).

Given,
$or, a - d + a + a +d = 12$

$or, 3a = 12$

$or, a = \frac{12}{3}$

$\therefore a = 4$


Also,
$(a -d)^2 + a^2 + (a +d)^2 = 66$

Using (a -b)² + (a +b)² = 2(a²+b²)

$or, 2(a^2 + d^2) + a^2 = 66$

$or, 2a^2 + 2d^2 + a^2 = 66$

$or, 3a^2 + 2d^2 = 66$

$or, 3(4)^2 + 2d^2 = 66$

$or, 2d^2 = 66 - 48$

$or, 2d^2 = 18$

$or, d^2 = \frac{18}{2}$

$or, d^2 = 9$

$or, d = \sqrt{9}$

$\therefore d = \pm 3$


When d = 3,
$(a -d) = (4-3) = 1$
$a = 4$
$(a + d) = (4+3) = 7$


When d = -3,
$(a -d) = (4+3) = 7$
$a = 4$
$(a +d) = (4-3) = 1$

Hence, the required numbers in Arithmetic Series are either 1,4,7 or 7,4,1.