If n Arithmetic means are inserted between two quantities a and b, then their sum is equal to


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Question Number 7522 by Little last updated on 02/Sep/16

$If n Arithmetic means are inserted between$ $two quantities a and b, then their sum is$ $equal to$
Commented by sandy_suhendra last updated on 02/Sep/16

$\frac{n + 2}{2} (a + b)$
Commented by Rohit last updated on 03/Sep/16

$\frac{a + c}{2} = b$
	Answered by Rasheed Soomro last updated on 02/Sep/16

	$L e t A_{1}, A_{2}, A_{3}, . . . A_{n} a r e n A M^{'} s b e t w e e n a a n d b .$ $a, A_{1}, A_{2}, A_{3}, . . . A_{n}, b a r e i n A P$ $L e t d b e c o m m o n f i f f r r n c e$ $a + (n + 2 - 1) d = (n + 2) t h t e r m = b$ $d = \frac{b - a}{n + 1}$ $A_{1} = a + \frac{b - a}{n + 1}, A_{2} = a + 2 (\frac{b - a}{n + 1}), A_{3} = a + 3 (\frac{b - a}{n + 1}), . . . .$ $A_{n} = a + n (\frac{b - a}{n + 1})$ $S = \frac{n}{2} [2 a + (n - 1) d]$ $S = \frac{n}{2} [2 A_{1} + (n - 1) d]$ $S = \frac{n}{2} [2 (a + \frac{b - a}{n + 1}) + (n - 1) (\frac{b - a}{n + 1})]$ $S = \frac{n}{2} [2 a + 2 (\frac{b - a}{n + 1}) + (n - 1) (\frac{b - a}{n + 1})]$ $S = \frac{n}{2} [2 a + (\frac{b - a}{n + 1}) [2 + n - 1]]$ $S = \frac{n}{2} [2 a + (\frac{b - a}{n + 1}) [n + 1]]$ $S = \frac{n}{2} [2 a + b - a]$ $S = \frac{n}{2} [a + b]$
	Commented by sandy_suhendra last updated on 03/Sep/16

	${w h a t}^{'} s t h e m e a n i n g o f t h e q u e s t i o n ?$ $A_{1} + A_{2} + . . . + A n o r a + A_{1} + A_{2} + . . . + A_{n} + b ?$ $S o r r y, m y E n g l i s h i s p o o r$ $i f a + A_{1} + . . . A_{n} + b$ $i n a, A_{1}, A_{2}, . . ., A n, b t h e r e a r e (n + 2) t e r m s$ $s o S = \frac{n}{2} [2 a + (n - 1) d] b e c o m e s S = \frac{n + 2}{2} [2 a + (n + 2 - 1) d]$
	Commented by Rasheed Soomro last updated on 04/Sep/16

	$M a y b e y o u a r e r i g h t . B u t$ $b y ‘ ‘ t h e i r {s u m}^{″} I u n d e r s t o o d$ $^{'} t h e s u m o f A M^{'} s^{'} h e r e .$
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