Question

find the three consecutive terms is an AP whose sum is 18 and the sum of their square is 140

Answer 1

let x+ x+1+ x+2=18
3x+3=18
x=15-3/3
x=12/3
x=4

now x=4
x+1=5
x+2=6

Answer 2

let the three no. be(a-d),a and(a+d).
according to question,
a+a-d+a+d=18
3a=18
a=6
a²+(a-d)²+(a+d)²=140
here,a=6
a²+a²+d²-2ad+a²+d²+2ad=140
3a²+2d²=140
3×6²+2d²= 140
2d²=140-108
2d²=32
d²=16
d=4
first term =a-d=6-4=2
second term=a=6
third term =a+d=10
AP:2,6,10,................
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