Question

are there not anyone to help me with this mad math problem?? If the sum of (n+1) terms of an arithmetic progression is an^2+bn+c then prove that a+c=b

Answer 1

sum of (n+1) term Sn+1 = an^2+bn+c

then S1= A1 = a (0^2)+b*0+c =c

& S2 = A1+A2 = a(1^2)+b*1+c =a+b+c
S3 = A1+A2+A3 = a(2^2)+b(2)+c =4a+2b+c


now S2-S1 =(A1+A2)-A1 =(a+b+c)-c

or A2 = a+b

S3 -S2 = (A1+A2+A3)-(A1+A2) =
A3 = (4a+2b+c)-(a+b+c) = 3a+b
now common difference

d = A3 -A2 = A2- A1

(3a+b) -(a+b)= (a+b)-c

2a = a+b-c

2a -a +c = b

a+c = b

proved