Question

If the first, second and the last terms of an A.P. are $a,b,c$ respectively, then the sum is

• A. $\frac{(a+b)(a+c-2b)}{2(b-a)}$
• B. $\frac{(b+c)(a+b-2c)}{2(b-a)}$
• C. $\frac{(a+c)(b+c-2a)}{2(b-a)}$
• D. None of these

Answer 1

The correct option is C $\frac{(a+c)(b+c-2a)}{2(b-a)}$

Given,
First term of AP $=a$
Second term of AP $=b$
Last term of AP $=c$

Then, Common difference $=b-a$
Using general equation of AP:
$c=a+(n-1)(b-a)$
$n=\frac{c+b-2a}{b-a}$

Then,
Sum of an AP $=\frac{n}{2}(2a+(n-1)\times d)$

$\frac{(c+b-2a)}{2(b-a)}(2\times a+(\frac{c+b-2a}{b-a}-1)\times (b-a\left)\right)$

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

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