(n) arithmetic means are inserted between (a) and (b).
The sequence is (a,\ A_1,\ A_2,\ \ldots,\ A_n,\ b) — total of (n+2) terms.
[ b = a + (n+1)d \implies d = \dfrac{b-a}{n+1} ]
Proved.
11TH•Mathematics•Sequences And Series•Question 5
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If $n$ arithmetic means are inserted between $a$ and $b$, prove that $d = \dfrac{b-a}{n+1}$, where $d$ is the common difference.
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