Question

Show that sum of an AP whose first term is a,the second term b and the last term c,is equal to (a+c)(b+c-2a)/2(b-a)..Plzzz i need hlpp!!!!

Answer 1

Given that the first term = a.

Second term = b

Last term = c

We know that common difference d = b - a

  Now,

Last term c = a + (n - 1) * d

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

                 c - a = (n - 1) * b - a

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

                 $n = 1 + \frac{c - a}{b - a}$

                $n = \frac{b - a + c - a}{b - a}$

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

  
Now,

Sum of n terms sn = n/2(a + c)

                        $= \frac{(b + c - 2a)}{2} (a + c)$
 
                        $\frac{(b + c - 21)(a + c)}{2(b-a)}$



Hope this helps!

Answer 2

plz refer to this attachment