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

first term = a,second term = b,last term = c
d= (b-a)
an = a + (n-1) d
=> c = a + (n-1) (b-a)
=> c = a + nb -an -b +a
=> c = 2a -an +nb -b
=> n = (b+c-2a) / (b-a)
Sn = n/2 (a+ l)
=> Sn = (b+c-2a) (a+c) / 2(b-a)

Answer 2

plz refer to this attachment