If tenth of an A.P. is 19 and sum of first fifteen
terms is 225 then fifth term of A.P. is

Options

A. 5
B. 6
C. 9
D. 7

Answer

To solve the problem step by step, we need to find the fifth term of an arithmetic progression (A.P.) given that the 10th term is 19 and the sum of the first 15 terms is 225.

### Step 1: Use the formula for the nth term of an A.P.
The nth term of an A.P. is given by the formula:
\[ T_n = a + (n - 1) \cdot d \]
where \(a\) is the first term and \(d\) is the common difference.

Given that the 10th term \(T_{10} = 19\), we can write:
\[ T_{10} = a + (10 - 1) \cdot d = a + 9d = 19 \quad \text{(Equation 1)} \]

### Step 2: Use the formula for the sum of the first n terms of an A.P.
The sum of the first n terms of an A.P. is given by the formula:
\[ S_n = \frac{n}{2} \cdot (2a + (n - 1) \cdot d) \]
Given that the sum of the first 15 terms \(S_{15} = 225\), we can write:
\[ S_{15} = \frac{15}{2} \cdot (2a + (15 - 1) \cdot d) = 225 \]
This simplifies to:
\[ \frac{15}{2} \cdot (2a + 14d) = 225 \]
Multiplying both sides by 2 to eliminate the fraction:
\[ 15 \cdot (2a + 14d) = 450 \]
Dividing both sides by 15:
\[ 2a + 14d = 30 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have two equations:
1. \(a + 9d = 19\) (Equation 1)
2. \(2a + 14d = 30\) (Equation 2)

From Equation 1, we can express \(a\) in terms of \(d\):
\[ a = 19 - 9d \]

Substituting \(a\) in Equation 2:
\[ 2(19 - 9d) + 14d = 30 \]
Expanding this:
\[ 38 - 18d + 14d = 30 \]
Combining like terms:
\[ 38 - 4d = 30 \]
Rearranging gives:
\[ -4d = 30 - 38 \]
\[ -4d = -8 \]
Dividing by -4:
\[ d = 2 \]

### Step 4: Find the first term \(a\)
Now substitute \(d = 2\) back into Equation 1 to find \(a\):
\[ a + 9(2) = 19 \]
\[ a + 18 = 19 \]
\[ a = 19 - 18 = 1 \]

### Step 5: Find the fifth term
Now we can find the fifth term \(T_5\) using the formula for the nth term:
\[ T_5 = a + (5 - 1) \cdot d = a + 4d \]
Substituting the values of \(a\) and \(d\):
\[ T_5 = 1 + 4 \cdot 2 = 1 + 8 = 9 \]

### Final Answer
The fifth term of the A.P. is **9**.

---