Sum of first n terms of an A.P is given by Sn = \(\frac{n}{2}\)(2a + (n − 1)d)
Given,
First term (a) = 5, last term (an) = 45 and sum of n terms (Sn) = 400
Now, we know that
an = a + (n – 1)d
⟹ 45 = 5 + (n – 1)d
⟹ 40 = nd – d
⟹ nd – d = 40 …. (1)
Also,
Sn = \(\frac{n}{2}\)(2(a) + (n − 1)d)
400 = \(\frac{n}{2}\)(2(5) + (n − 1)d)
800 = n (10 + nd – d)
800 = n (10 + 40) [using (1)]
⟹ n = 16
Putting n in (1), we find d
nd – d = 40
16d – d = 40
15d = 40
d = \(\frac{8}{3}\)
Therefore, the common difference of the given A.P. is \(\frac{8}{3}\).