The first and the last terms of an A.P. are 7 and 49 respectively. If

1.	The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find the common difference.
Answer» Given, First term (a) = 7, last term (a_n) = 49 and sum of n terms (S_n) = 420 Now, we know that a_n = a + (n – 1)d ⟹ 49 = 7 + (n – 1)d ⟹ 43 = nd – d ⟹ nd – d = 42 ….. (1) Next, S_n = \(\frac{n}{2}\)(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.

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

Answer»

Given,

First term (a) = 7, last term (a_n) = 49 and sum of n terms (S_n) = 420

Now, we know that

a_n = a + (n – 1)d

⟹ 49 = 7 + (n – 1)d

⟹ 43 = nd – d

⟹ nd – d = 42 ….. (1)

Next,

S_n = \(\frac{n}{2}\)(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.