Q. 154.7( 16 Votes )

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

Answer :

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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