Question

Prove that the sum of n arithmetic means between two numbers is
n times the single A.M between them

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Given that,

The sum of n arithmetic means between two numbers is  n times the single A.M between them.

Let a and b be two given number.

Let $A_{1}, A_{2}, A_{3}.....A_{n}$

So. $a, A_{1}, A_{2}, A_{3}......A_{n},b are in A.P</p><p><strong>We need to calculate the sum </strong></p><p><strong>Using formula of sum of A.P</strong></p><p>[tex]a+A_{1}+ A_{2}+A_{3}+.....+A_{n}+b=\dfrac{n+2}{2}(a+b)$...(I)

Here, first term = a

Last term = b

Number of terms = n+2

We need to prove that the n times the single A.M between a and b

Using equation (I)

$a+A_{1}+ A_{2}+A_{3}+.....+A_{n}+b=\dfrac{n+2}{2}(a+b)$

$A_{1}+ A_{2}+A_{3}+.....+A_{n}=\dfrac{(n+2)(a+b)}{2}-(a+b)$

$A_{1}+ A_{2}+A_{3}+.....+A_{n}=(a+b)(\dfrac{n+2}{2}-1)$

$A_{1}+ A_{2}+A_{3}+.....+A_{n}=(a+b)(\dfrac{n+2-2}{2})$

$A_{1}+ A_{2}+A_{3}+.....+A_{n}=(a+b)(\dfrac{n}{2})$

$A_{1}+ A_{2}+A_{3}+.....+A_{n}=n\dfrac{(a+b)}{2}$

Hence, The n times the single A.M between a and b its proved.