Given,
First term (a) = 7, last term (an) = 49 and sum of n terms (Sn) = 420
Now, we know that
an = a + (n – 1)d
⟹ 49 = 7 + (n – 1)d
⟹ 43 = nd – d
⟹ nd – d = 42 ….. (1)
Next,
Sn = \(\frac{n}{2}\)(2(7) + (n − 1)d)
⟹ 840 = n[14 + nd – d]
⟹ 840 = n[14 + 42] [using (1)]
⟹ 840 = 54n
⟹ n = 15 …. (2)
So, by substituting (2) in (1), we have
nd – d = 42
⟹ 15d – d = 42
⟹ 14d = 42
⟹ d = 3
Therefore, the common difference of the given A.P. is 3.