Determine the A.P whose third term is 16 and 7th term exceeds the 5th term by 12
[a] 2,4,6,8…
[b] 3,9,15,21,…
[c] 4,10,16,22,…
[d] 5,10,15,20,…

Hint: Assume that the first term of the A.P is a and the common difference is d. Using ${{a}_{n}}=a+\left( n-1 \right)d$ form two linear equations in two variable a and d.
Solve the system of equations using elimination method, or substitution method or graphically and hence determine a and d. Hence find the A.


Complete step-by-step solution -

Let the first term of the A.P be, a and the common difference be d.
We have ${{a}_{n}}=a+\left( n-1 \right)d$
Put n = 3, we get
$\begin{align}
  & {{a}_{3}}=a+2d=16 \\
 & \Rightarrow a+2d=16\text{ (i)} \\
\end{align}$
Put n = 7, we get
${{a}_{7}}=a+6d$
Put n = 5, we get
${{a}_{5}}=a+4d$
Hence we have $a+6d=a+4d+12$
Subtracting a+4d from both sides, we get
$2d=12$
Dividing both sides by 2, we get
$d=\dfrac{12}{2}=6$
Substituting the value of d in equation (i), we get
$a+2\left( 6 \right)=16$
Subtracting 12 from both sides, we get
$a=16-12=4$
Hence the A.P is 4,10,16,…
Hence option [c] is correct.

Note: [1] You can solve the question directly by eliminating the options. Among the given options only A.P which has the third term as 16 is in option [c]. Hence option [c] is the only correct answer.
[2] Difference between mth and nth term is given by ${{a}_{m}}-{{a}_{n}}=\left( m-n \right)d$
Hence if the difference between 7th term and 5th term is 12, we have
$\begin{align}
  & 12=\left( 7-5 \right)d \\
 & \Rightarrow d=6 \\
\end{align}$
which is the same as obtained above.