Question

The first term of an AP is $p$ and the common difference is $q$, then its 10th term is :

• A. $q - 9p$
• B. $p - 9q$
• C. $p + 9q$
• D. $2p + 9q$

Hint:

For any AP, the coefficient of $d$ in the $n$-th term is always $(n-1)$.
Thus, the 5th term is $a + 4d$, the 10th term is $a + 9d$, and the 100th term is $a + 99d$. This simple rule avoids calculation errors.

Answer 1

The Correct Option is C

Step 1: Build the sequence term by term.
Instead of substituting straight into the $n$th term formula, let us write out the arithmetic progression directly, one term at a time, using first term $p$ and common difference $q$.

Step 2: List the first several terms.
1st term: $p$
2nd term: $p + q$ (add the common difference once)
3rd term: $p + 2q$ (add the common difference twice)
4th term: $p + 3q$
5th term: $p + 4q$
Notice that the coefficient of $q$ in each term is always one less than the term's position number.

Step 3: Extend the pattern to the 10th term.
Continuing this counting:
6th term: $p + 5q$
7th term: $p + 6q$
8th term: $p + 7q$
9th term: $p + 8q$
10th term: $p + 9q$
For the 10th term, the common difference $q$ has been added exactly $10 - 1 = 9$ times to the first term.

Step 4: Write the final result.
\[ a_{10} = p + 9q \]
Final Answer:
The 10th term of the AP is $p + 9q$, which corresponds to option (C). \[ \boxed{p + 9q} \]