If 4th term of an A.p is zero, show that the 8th term is double the 6th term

To solve the problem, we need to use the properties of an arithmetic progression (A.P.). In an A.P., each term is generated by adding a constant difference to the previous term. Let's break down the steps to show that if the 4th term is zero, then the 8th term is double the 6th term.

Understanding the Terms of an A.P.

In an arithmetic progression, the nth term can be expressed as:

T_n = a + (n - 1)d

Here, a is the first term, d is the common difference, and n is the term number.

Finding the 4th Term

According to the problem, the 4th term (T₄) is zero:

T₄ = a + (4 - 1)d = a + 3d = 0

This equation tells us that:

a + 3d = 0

From this, we can express a in terms of d:

a = -3d

Calculating the 8th and 6th Terms

Now, let's find the 8th term (T₈) and the 6th term (T₆):

• For the 8th term:

T₈ = a + (8 - 1)d = a + 7d

Substituting the value of a:

T₈ = -3d + 7d = 4d

• For the 6th term:

T₆ = a + (6 - 1)d = a + 5d

Again, substituting the value of a:

T₆ = -3d + 5d = 2d

Establishing the Relationship

Now that we have both terms, we can compare them:

T₈ = 4d

T₆ = 2d

To show that the 8th term is double the 6th term, we can set up the following equation:

T₈ = 2 × T₆

Substituting the values we found:

4d = 2 × (2d)

Which simplifies to:

4d = 4d

This confirms that the 8th term is indeed double the 6th term, as required. Thus, we have shown that if the 4th term of an A.P. is zero, then the 8th term is double the 6th term.