An arithmetical progression has positive terms. The ratio of the differrence of the `4^(th)` and the `8^(th)` term to `15 ^(th)` term is `(4)/(15)` and the square of the difference of the `4^(th)` and the `1 ^(st)` term is 225. Which term of the series is 2015 ?

Answer

To solve the problem step by step, we will use the properties of an arithmetic progression (AP).

### Step 1: Define the terms of the AP
Let the first term of the arithmetic progression be \( a \) and the common difference be \( d \). The \( n^{th} \) term of the AP can be expressed as:
\[ a_n = a + (n-1)d \]

### Step 2: Write the expressions for the required terms
We need to find the 4th, 8th, and 15th terms:
- The 4th term \( a_4 = a + 3d \)
- The 8th term \( a_8 = a + 7d \)
- The 15th term \( a_{15} = a + 14d \)

### Step 3: Set up the first equation
According to the problem, the ratio of the difference of the 4th and 8th terms to the 15th term is \( \frac{4}{15} \):
\[ \frac{a_8 - a_4}{a_{15}} = \frac{4}{15} \]
Substituting the expressions we found:
\[ \frac{(a + 7d) - (a + 3d)}{a + 14d} = \frac{4}{15} \]
This simplifies to:
\[ \frac{4d}{a + 14d} = \frac{4}{15} \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives:
\[ 4d \cdot 15 = 4(a + 14d) \]
This simplifies to:
\[ 60d = 4a + 56d \]
Rearranging gives:
\[ 60d - 56d = 4a \]
\[ 4d = 4a \quad \Rightarrow \quad a = d \]

### Step 5: Set up the second equation
The problem states that the square of the difference of the 4th and 1st terms is 225:
\[ (a_4 - a_1)^2 = 225 \]
Substituting the terms:
\[ (a + 3d - a)^2 = 225 \]
This simplifies to:
\[ (3d)^2 = 225 \]
Taking the square root gives:
\[ 3d = 15 \quad \Rightarrow \quad d = 5 \]

### Step 6: Substitute back to find \( a \)
Since we found \( a = d \):
\[ a = 5 \]

### Step 7: Write the general term of the AP
Now we can write the general term of the AP:
\[ a_n = a + (n-1)d = 5 + (n-1) \cdot 5 = 5n \]

### Step 8: Find which term equals 2015
We need to find \( n \) such that:
\[ 5n = 2015 \]
Dividing both sides by 5:
\[ n = \frac{2015}{5} = 403 \]

### Final Answer
The term of the series that is equal to 2015 is the **403rd term**.

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