Question

Find the three consecutive terms in an $A.P.$ whose sum is $18$ and the sum of their squares is $140$.

Answer 1

Let $a-d,a,a+d$ be the three consecutive terms of an A.P.

Given$:$ Sum$=a-d+a+a+d=18$

$\Rightarrow 3a=18$ or $a=6$

Sum of their squares$={(6-d)}^2+6^2+{(6+d)}^2=140$

$\Rightarrow 36+d^2-12d+36+36+d^2+12d=140$

$\Rightarrow 2d^2=140-108=32$

$\Rightarrow d^2=16$

$\Rightarrow d=\pm 4$

The three consecutive terms are $6+4,4,6-4$ or $6-4,-4,6+4$

$\mathrm{10.4.2}$ or $2,-4,10$