Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.

Hint: An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……
a, (a+d), (a+2d), (a+3d),….. where a = first term, d = common difference.
a,b,c are said to be in AP if the common difference between any two consecutive number of the series is same ie \[b - a = c - b \Rightarrow 2b = a + c\]
Formula to consider for solving these questions
${a_n}=a+ (n− 1)d$
Where d -> common difference
A -> first term
n-> term
${a_n} -> {n^{th}}$ term

Complete step-by-step answer:
Let first term of AP = a
Let common difference of AP = d
It is given that its ${3^{rd}}$ term is equal to $16$.
Using formula , to find ${n^{th}}$ term of arithmetic progression,
\[\begin{array}{*{20}{l}}
  { \Rightarrow 16{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}\left( {3{\text{ }} - {\text{ }}1} \right){\text{ }}\left( d \right)} \\
  { \Rightarrow 16{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}2d \ldots \left( 1 \right)}
\end{array}\]
It is also given that the ${7^{th}}$ term exceeds ${5^{th}}$ term by $12$.
According to the given condition:
\[\begin{array}{*{20}{l}}
  { \Rightarrow a{\text{ }} + {\text{ }}(7-1){\text{ }}d{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}(5-1){\text{ }}d{\text{ }} + {\text{ }}12} \\

   \Rightarrow 2d{\text{ }} = {\text{ }}12 \\
   \Rightarrow d{\text{ }} = {\text{ }}6 \\

\end{array}\]
Putting value of d in equation 16 = a + 2d, from equation (1)
\[ \Rightarrow 16{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}2{\text{ }}\left( 6 \right) \Rightarrow a{\text{ }} = {\text{ }}4\]
Therefore, first term = a = 4
And, common difference = d = 6

Therefore, AP is 4, 10, 16, 22, 28, 34, 40, …

Note: To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a−d),a,(a+d)
4 terms: (a−3d),(a−d),(a+d),(a+3d)
5 terms: (a−2d),(a−d),a,(a+d),(a+2d)
${t_n}={S_n}-{S_{n-1}}$
If each term of an AP is increased, decreased, multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.
In an AP, the sum of terms equidistant from beginning and end will be constant.