Question

​The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference. 

Answer 1

Let a be the first term and d be the common difference.
We know that, sum of first n terms = S_n = \[\frac{n}{2}\][2a + (n − 1)d]
Also, n^th term = a_n = a + (n − 1)d
According to the question,
a = 7, a_n = 49 and S_n = 420

Now,
a_n = a + (n − 1)d
⇒ 49 = 7 + (n − 1)d
​⇒ 42 = nd − d
​⇒ nd − d = 42                     ....(1)

Also,
S_n = \[\frac{n}{2}\][2 × 7 + (n − 1)d]
⇒ 420 = \[\frac{n}{2}\][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.