# Prove that the ratio of sum of m arithmetic means between

Question:

Prove that the ratio of sum of m arithmetic means between the two numbers to the sum of n arithmetic means between them is m:n.

Solution:

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) $d=\frac{b-a}{n+1}$, where, $d$ is the common difference

n is the number of arithmetic means

(ii) $S_{n}=\frac{n}{2}[a+1]$ , Where n = Number of terms

$a=$ First term

$I=$ Last term

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

$a, A_{1}, A_{2}, A_{3} \ldots A_{m}, I$

In the above series we have (m + 2) terms

$\Rightarrow I=a+(m+2-1) d$

$\Rightarrow I=a+(m+1) d \ldots$ (i)

In the above series $A_{1}$ is second term

$\Rightarrow \mathrm{A}_{1}=\mathrm{a}+(2-1) \mathrm{d}$

$=\mathrm{a}+\mathrm{d}$

In the above series $A_{m}$ is the $(m+1)^{\text {th }}$ term

$\Rightarrow \mathrm{A}_{\mathrm{m}}=\mathrm{a}+(\mathrm{m}+1-1) \mathrm{d}$

$=\mathrm{a}+\mathrm{md}$

Now, $A_{1}+A_{m}=a+d+a+m d$

$=a+a+(m+1) d$

$=a+I$ [From eqn (i)]

Therefore, $A_{1}+A_{m}=a+1 \ldots$ (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, $S_{n}=\frac{n}{2}[a+1]$

$\mathrm{S}_{\mathrm{m}}=\frac{\mathrm{m}}{2}\left[\mathrm{~A}_{1}+\mathrm{A}_{\mathrm{m}}\right]$

From eqn. (ii)

$\mathrm{S}_{\mathrm{m}}=\frac{\mathrm{m}}{2}[\mathrm{a}+1]$

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

$a, A_{1}, A_{2}, A_{3} \ldots A_{n,} I$

In the above series we have (n + 2) terms

$\Rightarrow I=a+(n+2-1) d$

$\Rightarrow I=a+(n+1) d \ldots$ (ii)

In the above series $A_{1}$ is second term

$\Rightarrow \mathrm{A}_{1}=\mathrm{a}+(2-1) \mathrm{d}$

$=\mathrm{a}+\mathrm{d}$

In the above series $A_{n}$ is the $(n+1)^{\text {th }}$ term

$\Rightarrow \mathrm{A}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}+1-1) \mathrm{d}$

$=\mathrm{a}+\mathrm{nd}$

Now, $A_{1}+A_{n}=a+d+a+n d$

$=a+a+(n+1) d$

$=a+1$ [From eqn (iii)]

Therefore, $\mathrm{A}_{1}+\mathrm{A}_{\mathrm{n}}=\mathrm{a}+\mathrm{I} \ldots$ (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, $S_{n}=\frac{n}{2}[a+1]$

$S_{n}=\frac{n}{2}\left[A_{1}+A_{n}\right]$

From eqn. (iv)

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[\mathrm{a}+1]$

There, $\frac{S_{m}}{S_{n}}=\frac{\frac{m}{2}[a+1]}{\frac{n}{2}[a+1]}=\frac{m}{n}$

Hence Proved