Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73

Given, a11 = 38 and a16 = 73


We know that an = a + (n – 1)d


Hence, a11 = a + 10d = 38


And, a16 = a + 15d = 73


Subtracting 11th term from 16th term, we get following:


a + 15d – a – 10d = 73 – 38


Or, 5d = 35


Or, d = 7


Substituting the value of d in 11th term we get;


a + 10 x 7 = 38


Or, a + 70 = 38


Or, a = 38 – 70 = - 32


Now 31st term can be calculated as follows:


a31 = a + 30d


= - 32 + 30 x 7


= - 32 + 210 = 178


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