# 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 31st term.

So, we first find its 31st term.

Since, we know that

an = a + (n - 1) × d

a31 = 5 + (31 - 1) × 10

a31 = 5 + 30 × 10

a31 = 5 + 300

a31 = 305.

31st term of the AP is 305.

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

So, to find: place of the term 435.

So, let an = 435

Since, we know that

an = 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

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

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