Let `a_1,a_2,a_3, . . .,a_n` be in A.P. The ratio of sum of first five terms of the sum of first nine term is `5:17`. Also `110 lt a_15 lt 120`. Find the sum of first 10 terms of the A.P (where all `a_i (i=1,2,3, . . .,n)` are integers)

Options

A. 330
B. 460
C. 290
D. 380

Answer

To solve the problem step by step, we will follow the given information and derive the necessary equations to find the sum of the first 10 terms of the arithmetic progression (A.P.).

### Step 1: Understand the given information
We have an arithmetic progression (A.P.) represented by the terms \( a_1, a_2, a_3, \ldots, a_n \). The ratio of the sum of the first five terms to the sum of the first nine terms is given as \( \frac{5}{17} \). Additionally, the 15th term \( a_{15} \) lies between 110 and 120.

### Step 2: Write the formulas for the sums
The sum of the first \( n \) terms of an A.P. can be expressed as:
\[ S_n = \frac{n}{2} \times (2a + (n - 1)d) \]
where \( a \) is the first term and \( d \) is the common difference.

For the first 5 terms:
\[ S_5 = \frac{5}{2} \times (2a + 4d) = \frac{5}{2} \times (2a + 4d) \]

For the first 9 terms:
\[ S_9 = \frac{9}{2} \times (2a + 8d) = \frac{9}{2} \times (2a + 8d) \]

### Step 3: Set up the ratio equation
According to the problem, we have:
\[ \frac{S_5}{S_9} = \frac{5}{17} \]
Substituting the expressions for \( S_5 \) and \( S_9 \):
\[ \frac{\frac{5}{2} \times (2a + 4d)}{\frac{9}{2} \times (2a + 8d)} = \frac{5}{17} \]

### Step 4: Simplify the equation
Cancelling out \( \frac{5}{2} \) from both sides:
\[ \frac{2a + 4d}{9 \times (2a + 8d)} = \frac{5}{17} \]
Cross-multiplying gives:
\[ 17(2a + 4d) = 5 \times 9(2a + 8d) \]
Expanding both sides:
\[ 34a + 68d = 45(2a + 8d) \]
\[ 34a + 68d = 90a + 360d \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ 34a - 90a + 68d - 360d = 0 \]
\[ -56a - 292d = 0 \]
This simplifies to:
\[ 56a = -292d \quad \Rightarrow \quad 14a = -73d \quad \Rightarrow \quad d = -\frac{14}{73}a \]

### Step 6: Find \( a_{15} \)
The 15th term \( a_{15} \) is given by:
\[ a_{15} = a + 14d \]
Substituting \( d \):
\[ a_{15} = a + 14\left(-\frac{14}{73}a\right) = a - \frac{196}{73}a = a\left(1 - \frac{196}{73}\right) = a\left(\frac{73 - 196}{73}\right) = a\left(\frac{-123}{73}\right) \]
Given \( 110 < a_{15} < 120 \):
\[ 110 < -\frac{123}{73}a < 120 \]

### Step 7: Solve for \( a \)
Multiplying through by -73 (reversing the inequality):
\[ -110 \times 73 > 123a > -120 \times 73 \]
Calculating the bounds:
\[ -8030 > 123a > -8760 \]
Dividing by 123:
\[ \frac{-8030}{123} > a > \frac{-8760}{123} \]
Calculating:
\[ -65.2 > a > -71.2 \]
Since \( a \) must be a positive integer, we find that \( a \) can only be 2.

### Step 8: Calculate \( d \)
Using \( a = 2 \):
\[ d = -\frac{14}{73} \times 2 = -\frac{28}{73} \]

### Step 9: Find the sum of the first 10 terms
Using the sum formula:
\[ S_{10} = \frac{10}{2} \times (2a + 9d) = 5 \times (4 - \frac{252}{73}) = 5 \times \left(\frac{292 - 252}{73}\right) = 5 \times \frac{40}{73} = \frac{200}{73} \]

### Final Answer
The sum of the first 10 terms of the A.P. is:
\[ S_{10} = \frac{200}{73} \approx 2.74 \]