Answer:
Ok ok bommer whatever you say bommer
Step-by-step explanation:
We know, n
th
term T
n
=a+(n−1)d
a= First term
d= Common difference
n= number of terms
T
n
=n
th
term
Let the required AM be x
1
,x
2
,x
3
. Then,
6,x
1
,x
2
,x
3
,−6 are in AP.
Now, T
5
=−6
⇒a+4d=−6
⇒6+4d=−6
⇒4d=−12
⇒d=−3
∴x
1
=6+d=(6−3)=3,
x
2
=6+2d=(6−6)=0,
x
3
=6+3d=(6−9)=−3.
Three arithmetic means between 6 and −6 are 3,0,−3.
