If A be an arithmetic mean between two number | vedclass

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Question

(a) Let the two quantities be $a$ and $b$

and let ${A_1},\;{A_2}.......,{A_n}$ be the $n$ $A.M.'s$ between them.

Then $a,\;{A_1},\;{A_2}....

Vedclass · Accepted Answer

$S = n\,A$

Vedclass · Answer

(a) Let the two quantities be $a$ and $b$

and let ${A_1},\;{A_2}.......,{A_n}$ be the $n$ $A.M.'s$ between them.

Then $a,\;{A_1},\;{A_2}......{A_n},\;b$ are in $A.P. $ and let $d$ be the common difference.

Now ${T_{n + 2}} = b = a + (n + 2 - 1)d$

$\Rightarrow d = \frac{{b - a}}{{n + 1}}$

Also ${A_1} + {A_2} + ...... + {A_n} = {S_{n + 1}} - a$

$ = \frac{1}{2}(n + 1)\left[ {2a + (n + 1 - 1)\frac{{(b - a)}}{{(n + 1)}}} \right] - a$

= $\frac{n}{2}[2a + (b - a)] = \frac{n}{2}(a + b) = n\left( {\frac{{a + b}}{2}} \right) = nA$.

Trick: Let $1,3,5,7,9 $ is in $A.P.$

In this series $A = 5,n = 3,S = 15$

==> $S = nA$.

If $A$ be an arithmetic mean between two numbers and $S$ be the sum of $n$ arithmetic means between the same numbers, then

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