Question

Find three consecutive terms of an ap whose sum is 18 and sum of their squares is 140

Answer 1

Answer:

(x)+(x+1)=18

(x)^2+(x+1)^2=140

Answer 2

Answer: 2,6,10

Step-by-step explanation:

First term =a, common difference= d

3 consecutive terms in A.P is (a-d), a, (a+d)

Sum of terms=18

(a-d) +a+(a+d)=18

3a=18

a=6

Sum of their squares =140

(a-d) ^2 + a^2 +(a+d)^2=140

a^2+d^2-2ad+a^2+a^2+d^2+2ad=140

3a^2+2d^2=140

(3*6*6)+2d^2=140

108+2d^2=140

2d^2=140-108=32

d^2=32/2=16

d*d=16

d=4

3 terms are (a-d), a, (a+d)

(6-4), 6, (6+4)

2,6,10