Correct option is (D) b-a/n+1
Let a+d, a+2d, .........., a+nd are n arithmetic means inserted between a and b.
\(\therefore\) a, a+d, a+2d, .........., a+nd, b will form an A.P.
\(\therefore\) \(a+(n+1)d=b\)
\(\Rightarrow(n+1)d=b-a\)
\(\Rightarrow\) \(d=\frac{b-a}{n+1}\)
\(\therefore\) Common difference is \(d=\frac{b-a}{n+1}.\)