Question

the 4th and 10th term of an ap are 13 and 31 respectively. find the 21st 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 4th term \( a_4 = 13 \)
- The 10th term \( a_{10} = 31 \)

We can write the equations for these terms using the formula:

1. For the 4th term:
\[
a + 3d = 13 \quad \text{(1)}
\]

2. For the 10th term:
\[
a + 9d = 31 \quad \text{(2)}
\]

Now we can solve these two equations. We'll subtract equation (1) from equation (2):

\[
(a + 9d) - (a + 3d) = 31 - 13
\]

This simplifies to:

\[
6d = 18
\]

Dividing both sides by 6 gives:

\[
d = 3
\]

Now substituting \( d = 3 \) back into equation (1) to find \( a \):

\[
a + 3(3) = 13
\]

This simplifies to:

\[
a + 9 = 13
\]

So:

\[
a = 4
\]

Now that we have \( a = 4 \) and \( d = 3 \), we can find the 21st term using the formula:

\[
a_{21} = a + (21 - 1)d
\]
\[
a_{21} = 4 + 20(3)
\]
\[
a_{21} = 4 + 60
\]
\[
a_{21} = 64
\]

Thus, the 21st term of the arithmetic progression is:

\[
\boxed{64}
\]