Find the 31st term of an AP whose \(11^{th}\) term is 38 and \(16^{th}\) term is 73.

Given \(a_{11} = 38 \) and \(a_{16} = 73\)

Using formula \(a_n = a + ( n - 1)d \), to find nth term of arithmetic progression,

\(\Rightarrow \) 38 = a + (11 - 1)(d), and
73 = a + (16 - 1)(d)
\(\Rightarrow \) 38 = a + 10d, and
73 = a + 15d

These are equations consisting of two variables.

We have,
\(\Rightarrow \) 38 = a + 10d
\(\Rightarrow \) a = 38 - 10d
Let us put value of a in equation
\(\Rightarrow\) (73 = a + 15d),
\(\Rightarrow \) 73 = 38 - 10d + 15d
\(\Rightarrow \) 35 = 5d

Therefore, Common difference = d = 7

Putting value of d in equation
\(\Rightarrow \) 38 = a + 10d,
\(\Rightarrow \) 38 = a + 70
\(\Rightarrow \) a = -32

Therefore, common difference = d = 7 and
First term = a = –32

Using formula \(a_n = a + ( n - 1)d \), to find nth term of arithmetic progression,

\(\Rightarrow a_{31}\)= -32 + (31 - 1) (7)
\(\Rightarrow \) -32 + 210 = 178

Therefore, 31st term of AP is 178.