Let a be the first term and d be the common difference.
We know that, sum of first n terms
Sn = \(\frac n2\)[2a + (n-1)d]
Also, nth term = an = a + (n - 1)d
According to the question,
a = 7, an = 49 and Sn = 420
Now,
an = a + (n - 1)d
⇒ 49 = 7 + (n - 1)d
⇒ 42 = nd - d
⇒ nd - d = 42----(1)
Also,
Sn = \(\frac n2\)[2 x 7 + (n - 1)d]
⇒ 420 = \(\frac n2\)[14 + nd - d]
⇒ 840 = n[14 + 42] [from (1)]
⇒ 56n = 840
⇒ n = 15---(2)
On substituting (2) in (1), we get
nd − d = 42
⇒ (15 − 1)d = 42
⇒ 14d = 42
⇒ d = 3
Thus, common difference of the given A.P. is 3.