88
In an A.P., the sum of three consecutive terms is $24$ and the sum of their squares is $194$. Find the numbers.
Show SolutionHide Solution↓
Let the numbers be $a - d, a, a + d$
$\therefore a - d + a + a + d = 24$
$\Rightarrow a = 8$
Also, $(a - d)^2 + a^2 + (a + d)^2 = 194$
$\Rightarrow (8 - d)^2 + 8^2 + (8 + d)^2 = 194$
$\Rightarrow d^2 = 1 \Rightarrow d = \pm 1$
$\therefore$ Numbers are $7, 8, 9$ or $9,8,7$
$\therefore a - d + a + a + d = 24$
$\Rightarrow a = 8$
Also, $(a - d)^2 + a^2 + (a + d)^2 = 194$
$\Rightarrow (8 - d)^2 + 8^2 + (8 + d)^2 = 194$
$\Rightarrow d^2 = 1 \Rightarrow d = \pm 1$
$\therefore$ Numbers are $7, 8, 9$ or $9,8,7$