Solving Arithmetic Progressions: Proving Relationships Between Terms

Problem Statement and Background

Arithmetic progressions (APs) are sequences in which each term is obtained by adding a constant difference to the previous term. The n-th 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, and n is the term number.

Solving for the Relationship Between the 9th and 5th Terms

Step 1: Establishing the Relationship

Given that the 9th term is three times the 5th term, we can write this as:

[ a_9 3a_5 ]

By substituting the formula for the n-th term, we get:

[ a 8d 3(a 4d) ]

Expanding and simplifying the right side gives:

[ a 8d 3a 12d ]

Rearranging the equation:

[ a 8d - 3a - 12d 0 ]

This simplifies to:

[ -2a - 4d 0 ]

Dividing through by -2:

[ a 2d 0 ]

Solving for a:

[ a -2d ]

Step 2: Proving the Relationship Between the 8th and 4th Terms

Next, we need to prove that the 8th term is five times the 4th term, i.e.,

[ a_8 5a_4 ]

Using the formula for the terms, we have:

[ a_8 a 7d ] [ a_4 a 3d ]

Substituting a_4 into the equation for a_8:

[ a 7d 5(a 3d) ]

Expanding the right side:

[ a 7d 5a 15d ]

Rearranging gives:

[ a 7d - 5a - 15d 0 ]

This simplifies to:

[ -4a - 8d 0 ]

Dividing through by -4:

[ a 2d 0 ]

Given that we have already established that:

[ a 2d 0 ]

This confirms that:

[ a_8 5a_4 ]

Conclusion

The relationship between a and d is:

[ a -2d ]

And it has been proven that the 8th term of the AP is indeed five times the 4th term.