Question

The fourth and sixth term of an AP are 7 and 11 respectively. Find the ninth term

Answer 1

In an arithmetic progression (AP), the \( n \)-th term can be expressed as:

\[
a_n = a + (n-1)d
\]

where \( a \) is the first term and \( d \) is the common difference.

Given:
- The fourth term \( a_4 = 7 \):
\[
a + 3d = 7 \quad \text{(1)}
\]

- The sixth term \( a_6 = 11 \):
\[
a + 5d = 11 \quad \text{(2)}
\]

Now, we can solve these two equations simultaneously.

Let's subtract equation (1) from equation (2):

\[
(a + 5d) - (a + 3d) = 11 - 7
\]

This simplifies to:

\[
2d = 4
\]

Dividing both sides by 2 gives:

\[
d = 2
\]

Now, substitute \( d = 2 \) back into equation (1) to find \( a \):

\[
a + 3(2) = 7
\]
\[
a + 6 = 7
\]
\[
a = 1
\]

Now we have both \( a \) and \( d \):
- \( a = 1 \)
- \( d = 2 \)

We need to find the ninth term \( a_9 \):

\[
a_9 = a + 8d
\]
\[
a_9 = 1 + 8(2)
\]
\[
a_9 = 1 + 16 = 17
\]

Thus, the ninth term of the AP is \( \boxed{17} \).