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

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}, \dots 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}, \dots 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}), \dots .

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} + \dots + A n o r a + A_{1} + A_{2} + \dots + 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} + \dots A_{n} + b

i n a, A_{1}, A_{2}, \dots, 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 .