The 17th term of an AP exceeds its 10th term by 7. Find the common difference.
Step-by-Step Solution
Key Concept: Use the formula for the $n^{\text{th}}$ term of an arithmetic progression: $a_n = a + (n-1)d$, where $a$ is the first term and $d$ is the common difference. The difference between two terms of the same AP is the difference of their indices multiplied by $d$.
1. Let $a$ be the first term and $d$ be the common difference of the AP.
2. Write the expressions for the 17th and 10th terms using the $n^{\text{th}}$ term formula:
$$a_{17} = a + (17-1)d = a + 16d,$$
$$a_{10} = a + (10-1)d = a + 9d.$$
3. According to the question, the 17th term exceeds the 10th term by 7:
$$a_{17} - a_{10} = 7.$$
4. Substitute the expressions from step 2:
$$(a + 16d) - (a + 9d) = 7.$$
5. Simplify the left‑hand side:
$$a + 16d - a - 9d = 7 \[2mm] 7d = 7.$$
6. Solve for $d$:
$$d = \frac{7}{7} = 1.$$
7. Hence, the common difference of the AP is $\boxed{1}$.
Correct Answer: 1