Question

Which term of the AP 5, 15, 25, ….. will be 130 more than its 31^st term?

Answer 1

Here, a = 5 and d = (15 - 5) = 10

The 31^st 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, T_n = 435

\(\Rightarrow\) 5 + (n - 1) x 10 = 435

\(\Rightarrow\) 10n = 440

\(\Rightarrow\) n = 44

Hence, the 44^th term will be 130 more than its 31^st term