Let’s take a to be the first term and d to be the common difference.
And we know that, sum of first n terms
Sn = \(\frac{n}{2}\)(2a + (n − 1)d)
Given that sum of the first 7 terms of an A.P. is 63.
S7 = 63
And sum of next 7 terms is 161.
So, the sum of first 14 terms = Sum of first 7 terms + sum of next 7 terms
S14 = 63 + 161 = 224
Now, having
S7 = \(\frac{7}{2}\)(2a + (7 − 1)d)
⟹ 63(2) = 7(2a + 6d)
⟹ 9 × 2 = 2a + 6d
⟹ 2a + 6d = 18 . . . . (1)
And,
S14 = \(\frac{14}{2}\)(2a + (14 − 1)d)
⟹ 224 = 7(2a + 13d)
⟹ 32 = 2a + 13d …. (2)
Now, subtracting (1) from (2), we get
⟹ 13d – 6d = 32 – 18
⟹ 7d = 14
⟹ d = 2
Using d in (1), we have
2a + 6(2) = 18
2a = 18 – 12
a = 3
Thus, from nth term
⟹ a28 = a + (28 – 1)d
= 3 + 27 (2)
= 3 + 54 = 57
Therefore, the 28th term is 57.