If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term, find the AP.


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


According to question –


8th term of AP = a8 = 31 [Given]


a + (8 – 1)d = 31


a + 7d = 31…..(1)


15th term of AP = a15 = 16 + a11


a + (15 – 1)d = 16 + a + (11 – 1)d


14d = 16 + 10d


4d = 16


d = 4


Substituting the value of d in equation(1), we get –


a = 31 – 7 × 4 = 31 – 28 = 3


Thus, the required AP is 3,7,11,15,…….


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