The 17th term of an ap exceeds its 10th term by 7 find the common difference

General
1

the 17th term of an ap exceeds its 10th term by 7 find the common difference

2

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

Answer:
To determine the common difference of an arithmetic progression (AP) given that the 17th term exceeds the 10th term by 7, we can use the formula for the nth term of an AP. The nth term of an AP can be expressed as:

a_n = a + (n-1)d

where:

  • ( a ) is the first term,
  • ( d ) is the common difference,
  • ( n ) is the term number.

  1. Express the 17th term and the 10th term:

    For the 17th term (( a_{17} )):

    a_{17} = a + 16d

    For the 10th term (( a_{10} )):

    a_{10} = a + 9d

  2. Set up the given condition:

    According to the problem, the 17th term exceeds the 10th term by 7:

    a_{17} - a_{10} = 7

  3. Substitute the expressions for the 17th and 10th terms:

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

  4. Simplify the equation:

    a + 16d - a - 9d = 7
    7d = 7

  5. Solve for the common difference ( d ):

    d = \frac{7}{7} = 1

Therefore, the common difference ( d ) of the arithmetic progression is \boxed{1}.