Question

The nth term of an A.P. is \(3n + 2\). The common difference is :

• A. 8
• B. 2
• C. 5
• D. 3

Hint:

For any Arithmetic Progression whose \(n\)-th term is represented as a linear expression \(a_n = An + B\), the common difference is always equal to the coefficient of \(n\) (which is \(A\)).
In this case, since \(a_n = 3n + 2\), the coefficient of \(n\) is 3, so the common difference is immediately 3. This saves you from performing any calculation!

Answer 1

The Correct Option is D

Step 1: Understand what is being asked.
The $n$-th term of an A.P. is $a_n=3n+2$, and we need its common difference. Instead of computing two specific terms, let us prove the value of the common difference for a general term $n$ using the term right after it.

Step 2: Write the expression for the next term.
If the $n$-th term is $a_n=3n+2$, the term right after it, the $(n+1)$-th term, is found by replacing $n$ with $n+1$:
\[ a_{n+1} = 3(n+1)+2 = 3n+3+2 = 3n+5 \]
Step 3: Subtract to get the common difference.
By definition, the common difference $d$ is the term after minus the term before, and this value must be the same for every $n$ in an A.P:
\[ d = a_{n+1} - a_n = (3n+5)-(3n+2) \]
\[ d = 3n+5-3n-2 = 3 \]
Step 4: Note why this works.
Since $n$ cancelled out completely, the difference is 3 no matter which term we pick. This confirms 3 is the true common difference of this A.P, not just a value that happens to hold for the first two terms.

Final Answer:
The common difference of the A.P. is 3, matching option (D).
\[ \boxed{d=3} \]