Question

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

Answer 1

AP is 5, 15, 25,…

First term = a = 5

Common difference = d = 15 – 5 = 10

Find 31^st term:

a₃₁ = a + (n – 1)d

= 5 + (31 – 1) 10

= 5 + 30 x 10

= 305

Required term = 305 + 130 = 435

Now, say 435 be the nth term, then

a_n = a + (n – 1)d

435 = 5 + (n – 1)10

435 – 5 = (n – 1)10

n – 1 = 43

n = 44

The required term will be 44th term.