The eighth term of an AP is half its second term and the eleventh term exceeds one - third of its fourth term by 1. Find the 15th term.


Let the a be first term and d be common difference of AP


And we know that, that the nth term is


an = a + (n - 1)d


Given,



2a8 = a2


2(a + 7d) = a + d


2a + 14d = a + d


a = - 13d [ eqn1]


Also,



3(a + 10d) = a + 3d + 3


3a + 30d = a + 3d + 3


2a + 27d = 3


2 (- 13d) + 27d = 3 [ using eqn1]


d = 3


using this value in eqn1


a = - 13(3)


= - 39


Now,


a15 = a + 14d


= - 39 + 14(3)


= - 39 + 42


= 3


So 15th term is 3.


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