Question

Which term of the A.P. 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 × 10

= 305 

∴ Required term = (305 + 130) = 435

Let this be the n^th term.

Then, T_n = 435

`\implies` 5 + (n – 1) × 10 = 435 

`\implies` 10n = 440

`\implies` n = 44

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