the forth term of an A.P is 11 and the sixth term is 17 find the tenth term?

fourth term is 11 ---> a+3d = 11

sixth term is 17 ----> a+5d = 17
subtract them
2d = 6
d = 3
back in a+3d = 11
a + 9 = 11
a = 2

tenth term = a+9d = 2+9(3) = 29

a+9d =2+9(3)=29

To find the tenth term of an arithmetic progression (A.P), we can use the formula:

\(a_n = a_1 + (n-1)d\)

Where:
\(a_n\) is the \(n^{th}\) term,
\(a_1\) is the first term, and
\(d\) is the common difference.

Given that the fourth term (\(a_4\)) is 11 and the sixth term (\(a_6\)) is 17, we can use these values to find the common difference.

Step 1: Finding the common difference (d)
From the given data, we have:
\(a_4 = 11\) and \(a_6 = 17\)

Using the formula mentioned above, we can set up two equations:

\(11 = a_1 + 3d\) -- (Equation 1)
\(17 = a_1 + 5d\) -- (Equation 2)

Step 2: Solving for the common difference (d)
To eliminate \(a_1\) and find the value of \(d\), we can subtract Equation 1 from Equation 2:

\(17 - 11 = (a_1 + 5d) - (a_1 + 3d)\)

Simplifying, we get:
\(6 = 2d\)

Dividing both sides by 2, we find that \(d = 3\).

Step 3: Finding the tenth term (\(a_{10}\))
Using the common difference \(d = 3\) and any of the given terms (let's use \(a_4 = 11\)) in the formula, we can find the tenth term.

\(a_{10} = a_1 + (10 - 1)d\)

\(a_{10} = a_1 + 9d\)

\(a_{10} = 11 + 9(3)\)

\(a_{10} = 11 + 27\)

\(a_{10} = 38\)

Therefore, the tenth term of the arithmetic progression is 38.

To find the tenth term of an arithmetic progression (AP), where the fourth term is 11 and the sixth term is 17, we need to determine the common difference (d) of the AP.

Let's start by using the formula for the nth term of an AP:

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

where:
- \( a_n \) is the nth term of the AP.
- \( a \) is the first term of the AP.
- \( n \) is the position of the term.
- \( d \) is the common difference.

We are given that the fourth term (n=4) is 11, so we can substitute these values into the equation:

\[ a_4 = a + (4-1) \cdot d \]
\[ 11 = a + 3d \]

Similarly, we are given that the sixth term (n=6) is 17:

\[ a_6 = a + (6-1) \cdot d \]
\[ 17 = a + 5d \]

Now we have a system of equations:
\[ a + 3d = 11 \] (Equation 1)
\[ a + 5d = 17 \] (Equation 2)

To solve for the common difference (d), we need to subtract Equation 1 from Equation 2:

\[ (a + 5d) - (a + 3d) = 17 - 11 \]
\[ 5d - 3d = 6 \]
\[ 2d = 6 \]
\[ d = 3 \]

Now that we have the value of the common difference (d = 3), we can substitute it into Equation 1 or 2 to find the first term (a).

Using Equation 1:
\[ a + 3d = 11 \]
\[ a + 3(3) = 11 \]
\[ a + 9 = 11 \]
\[ a = 2 \]

So, the first term (a) is 2 and the common difference (d) is 3.

Finally, to find the tenth term (n = 10), we can substitute these values into the nth term formula:

\[ a_{10} = a + (10-1) \cdot d \]
\[ a_{10} = 2 + 9 \cdot 3 \]
\[ a_{10} = 2 + 27 \]
\[ a_{10} = 29 \]

Therefore, the tenth term of the arithmetic progression is 29.