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

In the given AP, the first term = a = 5

Common difference = d = 15 - 5 = 10

To find: place of the term which is 130 more than its 31^st term.

So, we first find its 31^st term.

Since, we know that

a_n = a + (n - 1) × d

∴ a₃₁ = 5 + (31 - 1) × 10

⇒ a₃₁ = 5 + 30 × 10

⇒ a₃₁ = 5 + 300

⇒ a₃₁ = 305.

∴ 31^st term of the AP is 305.

Now, 130 more than 31^st term of the AP is 130 + 305 = 435.

So, to find: place of the term 435.

So, let a_n = 435

Since, we know that

a_n = a + (n - 1) × d

∴ 435 = 5 + (n - 1) × 10

⇒ 435 - 5 = 10n - 10

⇒ 430 = 10n - 10

⇒ 430 + 10 = 10n

⇒ 10n = 440

⇒ n = 440/10 = 44

∴ 44^th term of the AP is the term which is 130 more than 31^st term.