Question:
Which term of the AP 5, 15, 25, ...will be 130 more than its 31st term?
Solution:
Here, a = 5 and d = (15 - 5) = 10
The 31st term is given by
T31 = a + (31 - 1)d = a + 30d = 5 + 30 ⨯ 10 = 305
∴ Required term = (305 + 130) = 435
Let this be be the nth term.
Then Tn = 435
⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
⇒ n = 44
Hence, the 44th term will be 130 more than its 31st term.