The correct option is D −2,−6,−10,−14
Let the mean be X1,X2,X3,X4 and the common difference be b; then 2,X1,X2,X3,X4,−18 are in A.P.;
⇒−18=2+5b⇒5b=−20⇒b=−4
Hence, X1=2+b=2+(−4)=−2;X2=2+2b=2−8=−6
X3=2+3b=2−12=−10;X4=2+4b=2−16=−14
The required means are −2,−6,−10,−14.