The 17th term of an AP exceeds its 10th term by 7. Find the common difference

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Question.

The 17th term of an AP exceeds its 10th term by 7. Find the common difference

Solution:

$a_{17}-a_{10}=7$

$(a+16 d)-(a+9 d)=7$

$7 d=7$

$d=1$

Therefore, the common difference is 1 .

The 17th term of an AP exceeds its 10th term by 7.