Question

show that the sum of an arithemetic series whose first term is A, second term is B, and the last term is C, is equal ( a+c) (b+c)-2a /(b-a(​

Answer 1

Answer:

$S_n= \frac{(b + c - 2a) \times (a + c)}{(b - a)}$

Step-by-step explanation:

First term = a

Second term = b

Last term = c

=> Common difference d = b - a

LET n be the number of terms

=> c = a + ( n - 1 )d

=> c = a + ( n - 1 )( b - a )

=> c - a = ( n - 1 )d

=> n - 1 = ( c - a )/( b - a )

=> n = ( c - a )/( b - a ) + 1

=> n = ( c - a + b - a )/( b - a )

=> n = ( b + c - 2a )/( b - a )

WKT

Sum of n terms of an AP with First term a and last term l is

Sn = (n/2)×(a+l)

Sn = ( b + c - 2a )/( b - a ) × ( a + c )

Sn = [( b + c - 2a ) × ( a + c )]/( b - a )

$\frac{(b + c - 2a) \times (a + c)}{(b - a)}$

HOPE IT HELPS