The 17th term of an AP exceeds its 10th term by 7.
Question.
The 17th term of an AP exceeds its 10th term by 7. Find the common difference
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 .
$a_{17}-a_{10}=7$
$(a+16 d)-(a+9 d)=7$
$7 d=7$
$d=1$
Therefore, the common difference is 1 .