Question

Pleass, solve it very urgent please with explainaitions

Answer 1

Answer:

Step-by-step explanation:

Substituting the value of s=n2(2l−(n−1)d), we get

4nd(2l−(n−1)d)=(d+2l)2

8ndl−(4n2−4n)d2=d2+4l2+4dl  

4l2+d2(4n2−4n+1)−4dl(2n−1)=0

(2l)2+(2nd−d)2−(2)(2l)(2nd−d)=0

(2l−2nd−d)2=0

Since squares of real numbers are always greater than or equal to 0

2l−d(2n−1)=0

So, d=2l2n−1 or in a much nicer form, d=2a where a is the first term of the AP.