Let first term of AP is a and common difference of AP is d.
We know that,
nth term of AP is an = a + (n – 1) d.
Given that,
8th term of AP is a8 = 31
⇒ a + (8 – 1) d = 31
⇒ a + 7d = 31 … (1)
Also,
Given that,
15th term of AP is 16 more than the 11th term of AP.
i.e., a15 = 16 + a11
⇒ a + 14d = 16 + a + 10d
⇒ 14d − 10d = 16
⇒ d = \(\frac{16}{4}\) = 4.
From equation (1),
a + 7 × 4 = 31
(∵d = 4)
⇒ a = 31 – 28 = 3.
Hence,
The first term of AP is a1 = a = 3.
The second term of AP is a2 = a + d
= 3 + 4 = 7.
The third term of AP is a3 = a + 2d
= 3 + 8
= 11.
Hence,
The required AP is 3, 7, 11, ……… .