To prove: ratio of sum of m arithmetic means between the two numbers to the
sum of n arithmetic means between them is m:n

a = First term
l = Last term
Let the first series of arithmetic mean having m arithmetic means be,
a, A1, A2, A3 … Am, l
In the above series we have (m + 2) terms
⇒ l = a + (m + 2 – 1)d
⇒ l = a + (m + 1)d … (i)
In the above series A1 is second term
⇒ A1 = a + (2-1)d
= a + d
In the above series Am is the (m+1)th term
⇒ Am = a + (m+1-1)d
= a + md
Now, A1 + Am = a + d + a + md
= a + a + (m+1)d

= a + l [From eqn (i)]
Therefore, A1 + Am = a + l … (ii)
For the sum of arithmetic means in the above series:-
First term = A1, Last term = Am, No. of terms = m

Let the second series of arithmetic mean having n arithmetic means be,
a, A1, A2, A3 … An, l
In the above series we have (n + 2) terms
⇒ l = a + (n + 2 – 1)d
⇒ l = a + (n + 1)d … (iii)
In the above series A1 is second term
⇒ A1 = a + (2-1)d
= a + d
In the above series An is the (n+1)th term
⇒ An = a + (n+1-1)d
= a + nd
Now, A1 + An = a + d + a + nd
= a + a + (n+1)d
= a + l [ From eqn (iii) ]

Therefore, A1 + An = a + l … (iv)
For the sum of arithmetic means in the above series:-
First term = A1, Last term = An, No. of terms = n

Hence Proved