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

Formula used: (i) , where, d is the common difference

n is the number of arithmetic means

(ii) , Where n = Number of terms

a = First term

l = Last term

Let the first series of arithmetic mean having m arithmetic means be,

a, A₁, A_2, A₃ … A_m, 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 A_m is the (m+1)^th term

⇒ A_m = a + (m+1-1)d

= a + md

Now, A₁ + A_m = a + d + a + md

= a + a + (m+1)d

= a + l [From eqn (i)]

Therefore, A₁ + A_m = a + l … (ii)

For the sum of arithmetic means in the above series:-

First term = A_1, Last term = A_m, No. of terms = m

Using Formula,

From eqn. (ii)

Let the second series of arithmetic mean having n arithmetic means be,

a, A₁, A_2, A₃ … A_n, 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 A_n is the (n+1)^th term

⇒ A_n = a + (n+1-1)d

= a + nd

Now, A₁ + A_n = a + d + a + nd

= a + a + (n+1)d

= a + l [ From eqn (iii) ]

Therefore, A₁ + A_n = a + l … (iv)

For the sum of arithmetic means in the above series:-

First term = A_1, Last term = A_n, No. of terms = n

Using Formula,

From eqn. (iv)

There,

Hence Proved