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.