Question

the 8th term of an ap is 5 times the third term while the 7th term is 9 greater than the 4th term .write the first five term of the Ap

Answer 1

To analyze the given information, let's denote the first term of the arithmetic progression (AP) as \( a \) and the common difference as \( d \).

The \( n \)-th term of an AP can be expressed as:
\[
T_n = a + (n-1)d
\]

**Given Information:**

1. The 8th term is 5 times the 3rd term:
\[
T_8 = 5 \cdot T_3
\]
Substituting the formula for the terms:
\[
a + 7d = 5(a + 2d)
\]

2. The 7th term is 9 greater than the 4th term:
\[
T_7 = T_4 + 9
\]
Substituting the formula for the terms:
\[
a + 6d = (a + 3d) + 9
\]

Now, we can solve these equations step by step.

### Step 1: Simplify the equations

**Equation 1:**
\[
a + 7d = 5(a + 2d)
\]
Expanding and rearranging:
\[
a + 7d = 5a + 10d
\]
\[
a - 5a + 7d - 10d = 0
\]
\[
-4a - 3d = 0 \quad \Rightarrow \quad 4a + 3d = 0 \quad \Rightarrow \quad 4a = -3d \quad \Rightarrow \quad a = -\frac{3d}{4}
\]

**Equation 2:**
\[
a + 6d = a + 3d + 9
\]
Subtracting \( a \):
\[
6d = 3d + 9
\]
\[
6d - 3d = 9
\]
\[
3d = 9 \quad \Rightarrow \quad d = 3
\]

### Step 2: Substitute \( d \) back to find \( a \)

Using the value of \( d \) to find \( a \):
\[
a = -\frac{3d}{4} = -\frac{3 \times 3}{4} = -\frac{9}{4}
\]

### Step 3: Determine the first five terms of the AP

Now that we have \( a \) and \( d \):
- \( a = -\frac{9}{4} \)
- \( d = 3 \)

We can now calculate the first five terms:
1. First term \( T_1 = a = -\frac{9}{4} \)
2. Second term \( T_2 = a + d = -\frac{9}{4} + 3 = -\frac{9}{4} + \frac{12}{4} = \frac{3}{4} \)
3. Third term \( T_3 = a + 2d = -\frac{9}{4} + 6 = -\frac{9}{4} + \frac{24}{4} = \frac{15}{4} \)
4. Fourth term \( T_4 = a + 3d = -\frac{9}{4} + 9 = -\frac{9}{4} + \frac{36}{4} = \frac{27}{4} \)
5. Fifth term \( T_5 = a + 4d = -\frac{9}{4} + 12 = -\frac{9}{4} + \frac{48}{4} = \frac{39}{4} \)

### Conclusion

The first five terms of the AP are:
\[
T_1 = -\frac{9}{4}, \quad T_2 = \frac{3}{4}, \quad T_3 = \frac{15}{4}, \quad T_4 = \frac{27}{4}, \quad T_5 = \frac{39}{4}
\]