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

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Solution

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 be the n^th term.
Then T_n = 435
⇒ 5 + (n - 1) ⨯ 10 = 435
⇒ 10n = 440
⇒ n = 44
Hence, the 44^th term will be 130 more than its 31^st term.

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