Here, a = 5 and d = (15 - 5) = 10
The 31st term is given by
T31 = a + (31 - 1) d = a + 30d = 5 + 30 x 10 = 305
∴ Required term = (305 + 130) = 435
Let this be the nth term.
Then, Tn = 435
\(\Rightarrow\) 5 + (n - 1) x 10 = 435
\(\Rightarrow\) 10n = 440
\(\Rightarrow\) n = 44
Hence, the 44th term will be 130 more than its 31st term