Let a be the first term and d be the common difference.
And we know that, sum of first n terms is:
Sn = \(\frac{n}{2}\)(2a + (n − 1)d)
Also, nth term is given by an = a + (n – 1)d
From the question, we have
Sq = 162 and a6 : a13 = 1 : 2
So,
2a6 = a13
⟹ 2 [a + (6 – 1d)] = a + (13 – 1)d
⟹ 2a + 10d = a + 12d
⟹ a = 2d …. (1)
And, S9 = 162
⟹ S9 = \(\frac{9}{2}\)(2a + (9 − 1)d)
⟹ 162 = \(\frac{9}{2}\)(2a + 8d)
⟹ 162 × 2 = 9[4d + 8d] [from (1)]
⟹ 324 = 9 × 12d
⟹ d = 3
⟹ a = 2(3) [from (1)]
⟹ a = 6
Hence, the first term of the A.P. is 6
For the 15th term, a15 = a + 14d = 6 + 14 × 3 = 6 + 42
Therefore, a15 = 48