Question

The sum of 3 consecutive terms of an A.P = 33 and the sum of their squares is equal to 371. Then the terms are 9,11, 13 or 13,11,9.

• A. True
• B. False

Answer 1

The correct option is A

True

Let the three numbers be a-d, a, a +d
(a-d) + a + (a+d) = 33
3a = 33
a = 11
and
${(a-d)}^2+a^2+{(a+d)}^2$ = $371$
${(11-d)}^2+a^2+{(11+d)}^2$ = $371$
$121+d^2-22d+121+121+d^2+22d$ = $371$
$363+2d^2$ = $371$
$2d^2$ = $8$
$d^2$ =$4$
$d=\pm 2$
$\therefore$ The numbers can be $(11-2),11,(11+2)$
9,11,13
or $(11+2),11,(11-2)$
13,11,9