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