The first , second and the last terms of an A.P. are `a ,b , c` respectively. Prove that the sum is `((a+c)(b+c)(c-2a))/(2(b-a))` .

Options

A. `(2(c-a))/(b-a)`
B. `(2c(c-a))/(b-a)+c`
C. `(2c(b-a))/(c-a)`
D. `(2b(c-a))/(b-a)`

Answer

B

Explanation

We have,
First term = a, Second term = b
`:." "`d=Common difference = b-a
It is given that the middle term is c. This means that there are an off number of terms in the A.P. Let there be (2n+1) terms in the A.P. Then, `(n+1)^(th)` term is middle term.
`:." "` Middle term = c
`rArr" "a+nd=crArra+n(b-a)=crArrn=(c-a)/(b-a)`
`:." Sum"=(2n+1)/(2){2a+(2n+1-1)d}`
`rArr" Sum"=(1)/(2){2((c-a)/(b-a))+1}{2a+2((c-a)/(b-a))(b-a)}`
`rArr" Sum"=(1)/(2){(2(c-a))/(b-a)+1}{2c}=(2c(c-a))/(b-a)+c`