Question

please answer it i will mark you as brainlist and follow ​

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

⇒3a=18 or a=6

Sum of their squares=(6−d)

2

+6

2

+(6+d)

2

=140

⇒36+d

2

−12d+36+36+d

2

+12d=140

⇒2d

2

=140−108=32

⇒d

2

=16

⇒d=±4

The three consecutive terms are 6+4,4,6−4 or 6−4,−4,6+4

10.4.2 or 2,−4,10