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If $A$ be an arithmetic mean between two numbers and $S$ be the sum of $n$ arithmetic means between the same numbers, then
$S = n\,A$
$A = n\,S$
$A = S$
None of these
Solution
(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 $A = 5,n = 3,S = 15$
==> $S = nA$.
