Determine the AP whose third term is 16 and the \(7^{th}\) term exceeds the \(5^{th}\) term by 12.

Let first term of AP = a
Let common difference of AP = d
It is given that its 3rd term is equal to 16.

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

\(\Rightarrow \) 16 = a + (3 - 1) (d)
\(\Rightarrow \) 16 = a + 2d… (1)

It is also given that \(7^{th}\) term exceeds \(5^{th}\) term by12.

According to the given condition:

\(\Rightarrow a_7 = a_5 + 12\)
\(\Rightarrow \) a + (7 – 1) d = a + (5 – 1) d + 12
\(\Rightarrow \) 2d = 12
\(\Rightarrow \) d = 6
Putting value of d in equation
\(\Rightarrow \) 16 = a + 2d,
\(\Rightarrow \) 16 = a + 2(6)
\(\Rightarrow \) a = 4

Therefore, first term = a = 4

And, common difference = d = 6

Therefore, AP is 4, 10, 16, 22…