Which term of the AP:3,15,27,39,..... will be 132 more than its 54th t

1.	Which term of the AP:3,15,27,39,..... will be 132 more than its 54th term?
Answer» GIVEN54th terma + 53 d132 more than a54a + 53d + 132Therefore, this is your an (last term)AP 3 15 27 39,...first term \'a\' = 3Common difference \'d\' = 15 - 3 = 12an = a + 53d + 132FORMULAan = a + (n - 1)da + 53d + 132 = a + (n - 1)dSubsitute the values of a and d3 + 53x12 + 132 = 3 + ( n - 1) x 123 + 53x12 + 132 - 3 = 12( n - 1)636 + 132 = 12( n - 1)768 = 12( n - 1)768 / 12 = n - 164 = n - 164 + 1 = nn = 65Therefore, the required term is 65th.

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